a. z = x² - 5y²: Predominantly hyperbolas.b. z = x² + 2y²: Predominantly ellipses.c. z = y - 3x²: Predominantly parabolas.d. z = -5x²: Predominantly lines.
To sketch the contour diagrams and determine the predominant shape of the contours for each function, we will plot a range of values for x and y and calculate the corresponding z-values.
a. z = x² - 5y²
Contour diagram:
```
| .
| .
| .
| .
| .
-----+-----------------
| .
| .
| .
| .
| .
```
The contour lines of this function are predominantly hyperbolas.
b. z = x² + 2y²
Contour diagram:
```
| .
| .
| .
| .
-----+-----------------
| .
| .
| .
|
|
```
The contour lines of this function are predominantly ellipses.
c. z = y - 3x²
Contour diagram:
```
| .
| .
| .
| .
-----+-----------------
| .
| .
| .
| .
|
```
The contour lines of this function are predominantly parabolas.
d. z = -5x²
Contour diagram:
```
| .
| .
| .
| .
-----+-----------------
|
|
|
|
|
```
The contour lines of this function are predominantly lines.
In summary:
a. z = x² - 5y²: Predominantly hyperbolas.
b. z = x² + 2y²: Predominantly ellipses.
c. z = y - 3x²: Predominantly parabolas.
d. z = -5x²: Predominantly lines.
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a. The contours of z = x² - 5y² are predominantly hyperbolas.
b. The contours of z = x² + 2y² are predominantly ellipses.
c. The contours of z = y - 3x² are predominantly parabolas.
d. The contours of z = -5x² are predominantly lines.
a. The function z = x² - 5y² represents contours that are predominantly hyperbolas. The contour lines are symmetric about the x-axis and y-axis, and they open up and down. The contours become closer together as they move away from the origin.
b. The function z = x² + 2y² represents contours that are predominantly ellipses. The contour lines are symmetric about the x-axis and y-axis, forming concentric ellipses centered at the origin. The contours become more elongated as they move away from the origin.
c. The function z = y - 3x² represents contours that are predominantly parabolas. The contour lines are symmetric about the y-axis, with each contour line being a vertical parabola. As the value of y increases, the parabolas shift upwards.
d. The function z = -5x² represents contours that are predominantly lines. The contour lines are straight lines parallel to the y-axis. Each contour line has a constant value of z, indicating that the function is a quadratic function with no dependence on y.
In summary, the contour diagrams for the given functions show that:
a. The contours of z = x² - 5y² are predominantly hyperbolas.
b. The contours of z = x² + 2y² are predominantly ellipses.
c. The contours of z = y - 3x² are predominantly parabolas.
d. The contours of z = -5x² are predominantly lines.
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Angela took a general aptitude test and scored in the 95 th percentile for aptitude in accounting. (a) What percentage of the scores were at or below her score? × % (b) What percentage were above? x %
The given problem states that Angela took a general aptitude test and scored in the 95th percentile for aptitude in accounting.
To find:(a) What percentage of the scores were at or below her score? × %(b) What percentage were above? x %
(a) The percentage of the scores that were at or below her score is 95%.(b) The percentage of the scores that were above her score is 5%.Therefore, the main answer is as follows:(a) 95%(b) 5%
Angela took a general aptitude test and scored in the 95th percentile for aptitude in accounting. (a) What percentage of the scores were at or below her score? × %(b) What percentage were above? x %The percentile score of Angela in accounting is 95, which means Angela is in the top 5% of the students who have taken the test.The percentile score determines the number of students who have scored below the candidate.
For example, if a candidate is in the 90th percentile, it means that 90% of the students who have taken the test have scored below the candidate, and the candidate is in the top 10% of the students. Therefore, to find out what percentage of students have scored below the Angela, we can subtract 95 from 100. So, 100 – 95 = 5. Therefore, 5% of the students have scored below Angela.
Hence, the answer to the first question is 95%.Similarly, to calculate what percentage of the students have scored above Angela, we need to take the value of the percentile score (i.e., 95) and subtract it from 100. So, 100 – 95 = 5. Therefore, 5% of the students have scored above Angela.
Thus, Angela's percentile score in accounting is 95, which means that she has scored better than 95% of the students who have taken the test. Further, 5% of the students have scored better than her.
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A boat is 80 miles away from the marina, sailing directly toward it at 20 miles per hour. Write an equation for the distance of the boat from the marina, d, after t hours.
If a boat is 80 miles away from the marina, sailing directly toward it at 20 miles per hour, then the equation for the distance of the boat from the marina, d, after t hours is d= 20t+ 80
To find the equation for the distance, follow these steps:
Assume the distance of the boat from the marina = d. After time t hours, the boat sails at 20 miles/hour, the direction is the same as the distance between boat and marina at time t. Therefore, the equation for the distance of the boat from the marina after t hours can be found by using the formula as follows: d = d₀ + vt, where,d₀ = initial distance between the boat and the marina = 80 miles, v = velocity of the boat = 20 miles/hour, t = time = t hours.Substituting these values, we get d = 80 + 20t ⇒d = 20t + 80.Learn more about distance:
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Compute ∂x^2sin(x+y)/∂y and ∂x^2sin(x+y)/∂x
The expression to be evaluated is `∂x²sin(x+y)/∂y` and `∂x²sin(x+y)/∂x`. The value of
`∂x²sin(x+y)/∂y = -cos(x+y)` and `
∂x²sin(x+y)/∂x = -cos(x+y)` respectively.
Compute ∂x²sin(x+y)/∂y
To begin, we evaluate `∂x²sin(x+y)/∂y` using the following formula:
`∂²u/∂y∂x = ∂/∂y (∂u/∂x)`.
The following are the differentiating processes:
`∂/∂x(sin(x+y)) = cos(x+y)`
The following are the differentiating processes:`
∂²(sin(x+y))/∂y² = -sin(x+y)
`Therefore, `∂x²sin(x+y)/∂y
= ∂/∂x(∂sin(x+y)/∂y)
= ∂/∂x(-sin(x+y))
= -cos(x+y)`
Compute ∂x²sin(x+y)/∂x
To begin, we evaluate
`∂x²sin(x+y)/∂x`
using the following formula:
`∂²u/∂x² = ∂/∂x (∂u/∂x)`.
The following are the differentiating processes:
`∂/∂x(sin(x+y)) = cos(x+y)`
The following are the differentiating processes:
`∂²(sin(x+y))/∂x²
= -sin(x+y)`
Therefore,
`∂x²sin(x+y)/∂x
= ∂/∂x(∂sin(x+y)/∂x)
= ∂/∂x(-sin(x+y))
= -cos(x+y)`
The value of
`∂x²sin(x+y)/∂y = -cos(x+y)` and `
∂x²sin(x+y)/∂x = -cos(x+y)` respectively.
Answer:
`∂x²sin(x+y)/∂y = -cos(x+y)` and
`∂x²sin(x+y)/∂x = -cos(x+y)`
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In a binary classification problem, based on k numeric features, describe a (hypothetical) situation where you expect a logistic regression to outperform linear discriminant analysis.
Logistic regression is expected to outperform linear discriminant analysis in a binary classification problem when there is a nonlinear relationship between the numeric features and the binary outcome.
Step 1: Consider a dataset with k numeric features and a binary outcome variable.
Step 2: Analyze the relationship between the numeric features and the binary outcome. If there is evidence of a nonlinear relationship, such as curved or non-monotonic patterns, logistic regression becomes advantageous.
Step 3: Fit logistic regression and linear discriminant analysis models to the dataset.
Step 4: Assess the performance of both models using appropriate evaluation metrics such as accuracy, precision, recall, or area under the receiver operating characteristic curve (AUC-ROC).
Step 5: Compare the performance of the logistic regression and linear discriminant analysis models. If logistic regression achieves higher accuracy, precision, recall, or AUC-ROC compared to linear discriminant analysis, it indicates that logistic regression outperforms linear discriminant analysis in capturing the nonlinear relationship between the features and the binary outcome.
In this hypothetical situation where there is a nonlinear relationship between the numeric features and the binary outcome, logistic regression is expected to outperform linear discriminant analysis by better capturing the complexity of the relationship and providing more accurate predictions.
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Using Frobenius method, obtain two linearly independent solutions
c. (1-x2)y"+2xy'+y=0 ans.
Y₁ = co (1- x²/ 2 +x4 + 8+...
Y2=C₁ x- x3/5+x5/40 + ...
Hint :r1= 1,r2 = 0
These two solutions, \(Y_1\) and \(Y_2\), are linearly independent because they cannot be written as scalar multiples of each other. Together, they form a basis for the general solution of the given differential equation.
The Frobenius method is used to find power series solutions to second-order linear differential equations. For the given equation, \(y'' + 2xy' + y = 0\), the Frobenius method yields two linearly independent solutions: \(Y_1\) and \(Y_2\).
The first solution, \(Y_1\), can be expressed as a power series: \(Y_1 = \sum_{n=0}^{\infty} c_nx^n\), where \(c_n\) are coefficients to be determined. Substituting this series into the differential equation and solving for the coefficients yields the series \(Y_1 = c_0(1 - \frac{x^2}{2} + x^4 + \ldots)\).
The second solution, \(Y_2\), is obtained by considering a different power series form: \(Y_2 = x^r\sum_{n=0}^{\infty}c_nx^n\). In this case, \(r = 0\) since it is given as one of the roots.
Substituting this form into the differential equation and solving for the coefficients gives the series \(Y_2 = c_1x - \frac{x^3}{5} + \frac{x^5}{40} + \ldots\).
These two solutions, \(Y_1\) and \(Y_2\), are linearly independent because they cannot be written as scalar multiples of each other. Together, they form a basis for the general solution of the given differential equation.
In the first solution, \(Y_1\), the terms of the power series represent the coefficients of successive powers of \(x\). By substituting this series into the differential equation,
we can determine the coefficients \(c_n\) by comparing the coefficients of like powers of \(x\). This allows us to find the values of the coefficients \(c_0, c_1, c_2, \ldots\), which determine the behavior of the solution \(Y_1\) near the origin.
The second solution, \(Y_2\), is obtained by considering a different power series form in which \(Y_2\) has a factor of \(x\) raised to the root \(r = 0\) multiplied by another power series. This form allows us to find a second linearly independent solution.
The coefficients \(c_n\) are determined by substituting the series into the differential equation and comparing coefficients. The resulting series for \(Y_2\) provides information about the behavior of the solution near \(x = 0\).
Together, the solutions \(Y_1\) and \(Y_2\) form a basis for the general solution of the given differential equation, allowing us to express any solution as a linear combination of these two solutions.
The Frobenius method provides a systematic way to find power series solutions and determine the coefficients, enabling the study of differential equations in the context of power series expansions.
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How many different 6-letter radio station call letters can be made
a. if the first letter must be G, W, T, or L and no letter may be repeated?
b. if repeats are allowed (but the first letter is G, W, T, or L)?
c. How many of the 6-letter radio station call letters (starting with G, W, T, or L) have no repeats and end with the letter H?
a. If the first letter must be G, W, T, or L and no letter may be repeated, there are 4 choices for the first letter and 25 choices for each subsequent letter (since repetition is not allowed). Therefore, the number of different 6-letter radio station call letters is 4 * 25 * 24 * 23 * 22 * 21.
b. If repeats are allowed (but the first letter is G, W, T, or L), there are still 4 choices for the first letter, but now there are 26 choices for each subsequent letter (including the possibility of repetition). Therefore, the number of different 6-letter radio station call letters is 4 * 26 * 26 * 26 * 26 * 26.
c. To find the number of 6-letter radio station call letters (starting with G, W, T, or L) with no repeats and ending with the letter H, we need to consider the positions of the letters. The first letter has 4 choices (G, W, T, or L), and the last letter must be H. The remaining 4 positions can be filled with the remaining 23 letters (excluding H and the first chosen letter). Therefore, the number of such call letters is 4 * 23 * 22 * 21 * 20.
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an airplane has crashed on a deserted island off the coast of fiji. the survivors are forced to learn new behaviors in order to adapt to the situation and each other.
In a case whereby the survivors are forced to learn new behaviors in order to adapt to the situation and each other. This is an example of Emergent norm theory.
What is Emergent norm?According to the emerging norm theory, groups of people congregate when a crisis causes them to reassess their preconceived notions of acceptable behavior and come up with new ones.
When a crowd gathers, neither a leader nor any specific norm for crowd conduct exist. Emerging conventions emerged on their own, such as the employment of umbrellas as a symbol of protest and as a defense against police pepper spray. To organize protests, new communication tools including encrypted messaging applications were created.
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complete question;
An airplane has crashed on a deserted island off the coast of Fiji. The survivors are forced to learn new behaviors in order to adapt to the situation and each other. This is an example of which theory?
at a hockey game, a vender sold a combined total of sodas and hot dogs. the number of sodas sold was more than the number of hot dogs sold. find the number of sodas sold and the number of hot dogs sold.
The selling was =
Number of sodas sold: 70
Number of hotdogs sold: 38
Given that a combined total of 108 sodas and hot dogs are sold at a game,
The number of hot dogs sold was 32 less than the number of sodas sold.
We need to find the number of each.
Let's denote the number of sodas sold as "S" and the number of hot dogs sold as "H".
We know that the combined total of sodas and hot dogs sold is 108, so we can write the equation:
S + H = 108
We're also given that the number of hot dogs sold is 32 less than the number of sodas sold.
In equation form, this can be expressed as:
H = S - 32
Now we can substitute the second equation into the first equation:
S + (S - 32) = 108
Combining like terms:
2S - 32 = 108
Adding 32 to both sides:
2S = 140
Dividing both sides by 2:
S = 70
So the number of sodas sold is 70.
To find the number of hot dogs sold, we can substitute the value of S into one of the original equations:
H = S - 32
H = 70 - 32
H = 38
Therefore, the number of hot dogs sold is 38.
To summarize:
Number of sodas sold: 70
Number of hotdogs sold: 38
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Complete question =
At a hockey game, a vender sold a combined total of 108 sodas and hot dogs. The number of hot dogs sold was 32 less than the number of sodas sold. Find the number of sodas sold and the number of hot dogs sold.
NUMBER OF SODAS SOLD:
NUMBER OF HOT DOGS SOLD:
The language Balanced over Σ={(,), } is defined recursively as follows 1. Λ∈ Balanced. 2. ∀x,y∈ Balanced, both xy and (x) are elements of Balanced. A prefix of a string x is a substring of x that occurs at the beginning of x. Prove by induction that a string x belongs to this language if and only if (iff) the statement B(x) is true. B(x) : x contains equal numbers of left and right parentheses, and no prefix of x contains more right than left. Reminder for this and all following assignments: if you need to prove the "iff" statement, i.e., X⟺ Y, you need to prove both directions, namely, "given X, prove that Y follows from X(X⟹Y) ", and "given Y, prove that X follows from Y(X⟸Y) ".
The language Balanced over Σ = {(, )} is defined recursively as follows: Λ ∈ Balanced, and ∀ x, y ∈ Balanced, both xy and (x) are elements of Balanced. To prove by induction that a string x belongs to this language if and only if the statement B(x) is true. B(x): x contains equal numbers of left and right parentheses, and no prefix of x contains more right than left.
The induction proof can be broken down into two parts as follows: (X ⟹ Y) and (Y ⟹ X).
Let's start by proving that (X ⟹ Y):
Base case: Λ ∈ Balanced. The statement B(Λ) is true since it contains no parentheses. Therefore, the base case holds.
Inductive case: Let x ∈ Balanced and suppose that B(x) is true. We must show that B(xy) and B(x) are both true.
Case 1: xy is a balanced string. xy has equal numbers of left and right parentheses. Thus, B(xy) is true.
Case 2: xy is not balanced. Since x is balanced, it must contain equal numbers of left and right parentheses. Therefore, the number of left parentheses in x is greater than or equal to the number of right parentheses. If xy is not balanced, then it must have more right parentheses than left. Since all of the right parentheses in xy come from y, y must have more right than left. Thus, no prefix of y contains more left than right. Therefore, B(x) is true in this case. Thus, the inductive case holds and (X ⟹ Y) is true.
Now let's prove that (Y ⟹ X):
Base case: Λ has equal numbers of left and right parentheses, and no prefix of Λ contains more right than left. Since Λ contains no parentheses, both statements hold. Therefore, the base case holds.
Inductive case: Let x be a string with equal numbers of left and right parentheses, and no prefix of x contains more right than left. We must show that x belongs to this language. We can prove this by showing that x can be constructed using the two rules that define the language. If x contains no parentheses, it is equal to Λ, which belongs to the language. Otherwise, we can write x as (y) or xy, where y and x are both balanced strings. Since y is a substring of x, it follows that no prefix of y contains more right than left. Also, y contains equal numbers of left and right parentheses. Thus, by induction, y belongs to the language. Similarly, since x is a substring of xy, it follows that x contains equal numbers of left and right parentheses. Moreover, x contains no more right parentheses than left because y, which has no more right than left, is a substring of xy. Thus, by induction, x belongs to the language. Therefore, the inductive case holds, and (Y ⟹ X) is true.
In conclusion, since both (X ⟹ Y) and (Y ⟹ X) are true, we can conclude that x belongs to this language if and only if B(x) is true.
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Prove that the maximum number of edges in a bipartite subgraph of the Petersen graph is ≤13. (b) Find a bipartite subgraph of the Petersen graph with 12 edges.
(a) Maximum edges in bipartite subgraph of Petersen graph ≤ 13.
(b) Example bipartite subgraph of Petersen graph with 12 edges.
(a) To prove that the maximum number of edges in a bipartite subgraph of the Petersen graph is ≤13, we can use the fact that the Petersen graph has 10 vertices and 15 edges.
Assume that we have a bipartite subgraph of the Petersen graph. Since it is bipartite, we can divide the 10 vertices into two sets, A and B, such that all edges in the subgraph are between vertices from set A and set B.
Now, let's consider the maximum number of edges that can exist between the two sets, A and B. The maximum number of edges will occur when all vertices from set A are connected to all vertices from set B.
In the Petersen graph, each vertex is connected to exactly three other vertices. Therefore, in the bipartite subgraph, each vertex in set A can have at most three edges connecting it to vertices in set B. Since set A has 5 vertices, the maximum number of edges from set A to set B is 5 * 3 = 15.
Similarly, each vertex in set B can have at most three edges connecting it to vertices in set A. Since set B also has 5 vertices, the maximum number of edges from set B to set A is also 5 * 3 = 15.
However, each edge is counted twice (once from set A to set B and once from set B to set A), so we need to divide the total count by 2. Therefore, the maximum number of edges in the bipartite subgraph is 15 / 2 = 7.5, which is less than or equal to 13.
Hence, the maximum number of edges in a bipartite subgraph of the Petersen graph is ≤13.
(b) To find a bipartite subgraph of the Petersen graph with 12 edges, we can divide the 10 vertices into two sets, A and B, such that each vertex in set A is connected to exactly two vertices in set B.
One possible bipartite subgraph with 12 edges can be formed by choosing the following sets:
- Set A: {1, 2, 3, 4, 5}
- Set B: {6, 7, 8, 9, 10}
In this subgraph, each vertex in set A is connected to exactly two vertices in set B, resulting in a total of 10 edges. Additionally, we can choose two more edges from the remaining edges of the Petersen graph to make a total of 12 edges.
Note that there may be other valid bipartite subgraphs with 12 edges, but this is one example.
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Assume fand g are differentiable functions with h(x)=f(g(x)) Suppose the equation of the line langent to the graph of g at the point (3,6) is y=4x−6 and the equation of the line tangent to the graph of f at (6,8) is y=2x−4 a. Calculate h(3) and h'(3) b. Determine an equation of the line tangent to the graph of h at the point on the graph where x=3.
The equation of the line tangent to h at the point where [tex]x = 3[/tex] is [tex]y - h(3) = 8(x - 3).[/tex]
b. Determine an equation of the line tangent to the graph of h at the point on the graph where x = 3.
Using Chain Rule, [tex]$\frac{dh}{dx}=f'(g(x)) \cdot g'(x)$[/tex]
Therefore,
$[tex]\frac{dh}{dx}\Bigg|_{x=3}\\=f'(g(3)) \cdot g'(3)\\=f'(6) \cdot 4\\=\\2 \cdot 4 \\=8$[/tex]
Therefore, at x = 3, the slope of the tangent line to h is 8.
Also, we know that (3, h(3)) lies on the tangent line to h at x = 3.
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verify that each given function is a solution of the differential equation. 5. y"-y=0; y_1(t) = e', y_2(t) = cosh t
This equation is not satisfied for all values of t, so y_2(t) = cosh(t) is not a solution of the differential equation y'' - y = 0.
To verify that y_1(t) = e^t is a solution of the differential equation y'' - y = 0, we need to take the second derivative of y_1 and substitute both y_1 and its second derivative into the differential equation:
y_1(t) = e^t
y_1''(t) = e^t
Substituting these into the differential equation, we get:
y_1''(t) - y_1(t) = e^t - e^t = 0
Therefore, y_1(t) = e^t is indeed a solution of the differential equation.
To verify that y_2(t) = cosh(t) is also a solution of the differential equation y'' - y = 0, we follow the same process:
y_2(t) = cosh(t)
y_2''(t) = sinh(t)
Substituting these into the differential equation, we get:
y_2''(t) - y_2(t) = sinh(t) - cosh(t) = 0
This equation is not satisfied for all values of t, so y_2(t) = cosh(t) is not a solution of the differential equation y'' - y = 0.
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MODELING WITH MATHEMATICS The function y=3.5x+2.8 represents the cost y (in dollars ) of a taxi ride of x miles. a. Identify the independent and dependent variables. b. You have enough money to travel at most 20 miles in the taxi. Find the domain and range of the function.
a. The independent variable is x (number of miles traveled) and the dependent variable is y (cost of the taxi ride).
b. The domain of the function is x ≤ 20 (maximum distance allowed) and the range is y ≤ 72.8 (maximum cost for a 20-mile ride).
a. The independent variable is x, representing the number of miles traveled in the taxi. The dependent variable is y, representing the cost of the taxi ride in dollars.
b. The given function is y = 3.5x + 2.8, which represents the cost of a taxi ride based on the number of miles traveled. To find the domain and range of the function for a maximum distance of 20 miles, we need to consider the possible values for x and y within that range.
Domain:
Since the maximum distance allowed is 20 miles, the domain of the function is the set of all possible x-values that satisfy this condition. Therefore, the domain of the function is x ≤ 20.
Range:
To determine the range, we need to calculate the possible values for y corresponding to the given domain. Plugging in the maximum distance of 20 miles into the function, we have:
y = 3.5(20) + 2.8
y = 70 + 2.8
y = 72.8
Hence, the range of the function for a maximum distance of 20 miles is y ≤ 72.8.
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What is the asymptotic relationship between x and x2(2+sin(x)) Select all that apply x=O(x2(2+sin(x)))x=Θ(x2(2+sin(x)))x=Ω(x2(2+sin(x)))x=ω(x2(2+sin(x)))x=o(x2(2+sin(x)))
Expert Answer
The asymptotic relationship between x and x^2(2+sin(x)) is x=Θ(x^2(2+sin(x))) and x=o(x^2(2+sin(x))).
To determine the asymptotic relationship between x and x^2(2+sin(x)), we need to examine the growth rates of these functions as x approaches infinity.
x^2(2+sin(x)) grows faster than x because the x^2 term dominates over x. Additionally, the sinusoidal term sin(x) does not affect the overall growth rate significantly as x becomes large.
Based on this analysis, we can conclude the following relationships:
x=Θ(x^2(2+sin(x))): This notation indicates that x and x^2(2+sin(x)) have the same growth rate. As x approaches infinity, the difference between the two functions becomes negligible.
x=o(x^2(2+sin(x))): This notation indicates that x grows at a slower rate than x^2(2+sin(x)). In other words, the growth of x is "smaller" compared to x^2(2+sin(x)) as x becomes large.
Other notations such as x=O(x^2(2+sin(x))), x=Ω(x^2(2+sin(x))), and x=ω(x^2(2+sin(x))) do not accurately represent the relationship between x and x^2(2+sin(x)). These notations imply upper or lower bounds on the growth rates, but they do not capture the precise relationship between the two functions.
In summary, the correct asymptotic relationships are x=Θ(x^2(2+sin(x))) and x=o(x^2(2+sin(x))).
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Guided Practice: Problem 1 The amount of memory available on an iPhone seems like it doubles with each new version. If this is true, and the first version had 4 gigabytes of memory, how much memory does the 10^(th) version have?
The 10th version of the iPhone would have 4096 gigabytes (or 4 terabytes) of memory.
If the amount of memory on an iPhone doubles with each new version, we can use exponential growth to find the amount of memory for the 10th version.
Given that the first version had 4 gigabytes of memory, we can express the amount of memory for each version as a power of 2. Let's denote the amount of memory for the nth version as M(n).
We can see that M(1) = 4 gigabytes.
Since each new version doubles the memory, we can express M(n) in terms of M(n-1) as follows:
M(n) = 2 * M(n-1)
Using this recursive formula, we can calculate the amount of memory for the 10th version:
M(10) = 2 * M(9)
= 2 * (2 * M(8))
= 2 * (2 * (2 * M(7)))
= 2 * (2 * (2 * (2 * (2 * (2 * (2 * (2 * (2 * M(1)))))))))
Substituting M(1) = 4, we can simplify the expression:
M(10) = 2^10 * M(1)
= 2^10 * 4
= 1024 * 4
= 4096
Therefore, the 10th version of the iPhone would have 4096 gigabytes (or 4 terabytes) of memory.
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Find the equation of the tangent line to y=8e^x
at x=8. (Use symbolic notation and fractions where needed.) y= Incorrect Try to guess a formula for f ′ (x) where f(x)=2x.f ′(x)=
The equation of the tangent line to the curve [tex]y = 8e^x[/tex] at x = 8 is given by [tex]y - 8e^8 = 8 * e^8 (x - 8).[/tex]
To find the equation of the tangent line to the curve [tex]y = 8e^x[/tex] at x = 8, we first need to find the derivative of the function [tex]y = 8e^x.[/tex]
Let's differentiate [tex]y = 8e^x[/tex] with respect to x:
[tex]d/dx (y) = d/dx (8e^x)[/tex]
Using the chain rule, we have:
[tex]dy/dx = 8 * d/dx (e^x)[/tex]
The derivative of [tex]e^x[/tex] with respect to x is simply [tex]e^x[/tex]. Therefore:
[tex]dy/dx = 8 * e^x[/tex]
Now, we can find the slope of the tangent line at x = 8 by evaluating the derivative at that point:
slope = dy/dx at x
= 8
[tex]= 8 * e^8[/tex]
To find the equation of the tangent line, we use the point-slope form:
y - y1 = m(x - x1)
Where (x1, y1) represents the point on the curve where the tangent line touches, and m is the slope.
In this case, x1 = 8, [tex]y_1 = 8e^8[/tex], and [tex]m = 8 * e^8[/tex]. Plugging these values into the equation, we get:
[tex]y - 8e^8 = 8 * e^8 (x - 8)[/tex]
This is the equation of the tangent line to the curve [tex]y = 8e^x[/tex] at x = 8.
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Use pumping Lemma to prove that the following languages are not regular L3={ωωRβ∣ω,β∈{0,1}+} . L4={1i0j1k∣i>j and i0}
The language L3 is not regular. It can be proven using the pumping lemma for regular languages.
Here is the proof:
Assume L3 is a regular language.
Let w = xyβ, where β is a non-empty suffix of ω and x is a prefix of ω of length p or greater.
We can write w as w = xyβ = ωαββ R, where α is the suffix of x of length p or greater. Because L3 is a regular language, there exists a string v such that uviw is also in L3 for every i ≥ 0.
Let i = 0.
Then u0viw = ωαββR is in L3. By the pumping lemma, we have that v = yz and |y| > 0 and |uvyz| ≤ p. But this means that we can pump y any number of times and still get a string in L3, which is a contradiction.
Therefore, L3 is not a regular language.
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Find the area of the surface obtained by rotating the curve x=8 cos ^{3} θ, y=8 sin ^{3} θ, 0 ≤ θ ≤ π / 2 about the y -axis.
The area of the surface obtained by rotating the curve x = 8 cos³(θ), y = 8 sin³(θ), 0 ≤ θ ≤ π/2, about the y-axis is 32π/3 square units.
How did we get the value?To find the area of the surface obtained by rotating the curve about the y-axis, we can use the formula for surface area of revolution. The formula is given by:
A = 2π∫[a, b] x × √(1 + (dx/dy)²) dy,
where [a, b] is the interval of integration along the y-axis.
Let's start by finding the expression for dx/dy:
x = 8 cos³(θ)
dx/dθ = -24 cos²(θ)sin(θ)
dx/dy = (dx/dθ) / (dy/dθ)
y = 8 sin³(θ)
dy/dθ = 24 sin²(θ)cos(θ)
dx/dy = (-24 cos²(θ)sin(θ)) / (24 sin²(θ)cos(θ))
= - cos(θ) / sin(θ)
= -cot(θ)
Now, we need to determine the interval of integration, [a, b], which corresponds to the given range of θ, 0 ≤ θ ≤ π/2. In this range, sin(θ) and cos(θ) are always positive, so we can express the interval as [0, π/2].
Using the given information, the formula for the surface area of revolution becomes:
A = 2π∫[0, π/2] (8 cos³(θ)) × √(1 + (-cot(θ))²) dy
= 16π∫[0, π/2] cos³(θ) × √(1 + cot²(θ)) dy
To simplify the integral, we can use the trigonometric identity: 1 + cot²(θ) = csc²(θ).
A = 16π∫[0, π/2] cos³(θ) × √(csc²(θ)) dy
= 16π∫[0, π/2] cos³(θ) × csc(θ) dy
Now, let's proceed with the integration:
A = 16π∫[0, π/2] (cos³(θ) / sin(θ)) dy
To simplify further, we can express the integral in terms of θ instead of y:
A = 16π∫[0, π/2] (cos³(θ) / sin(θ)) (dy/dθ) dθ
= 16π∫[0, π/2] cos³(θ) dθ
Now, we need to evaluate this integral:
A = 16π∫[0, π/2] cos³(θ) dθ
This integral can be solved using various methods, such as integration by parts or trigonometric identities. Let's use the reduction formula to evaluate it:
[tex]∫ cos^n(θ) dθ = (1/n) × cos^(n-1)(θ) × sin(θ) + [(n-1)/n] × ∫ cos^(n-2)(θ) dθ[/tex]
Applying this formula to our integral, we have:
[tex]A = 16π * [(1/3) * cos^2(θ) * sin(θ) + (2/3) * ∫ cos(θ) dθ]\\= 16π * [(1/3) * cos^2(θ) * sin(θ) + (2/3) * sin(θ)]
[/tex]
Now, let's evaluate the definite integral
for the given range [0, π/2]:
[tex]A = 16π * [(1/3) * cos^2(π/2) * sin(π/2) + (2/3) * sin(π/2)] \\= 16π * [(1/3) * 0 * 1 + (2/3) * 1]\\= 16π * (2/3)\\= 32π/3[/tex]
Therefore, the area of the surface obtained by rotating the curve x = 8 cos³(θ), y = 8 sin³(θ), 0 ≤ θ ≤ π/2, about the y-axis is 32π/3 square units.
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Assume that on a camping trip, the probability of being attacked by a bear is P=0.25×10 −6. If a camper goes camping 20 times a year, what is the probability of being attacked by a bear within the next 20 years? (Assume that the trips are independent.)
The probability of at least 1 attack in 20 years is approximately:
We can solve this problem by using the binomial distribution formula, where:
n = number of trials = 20 years
p = probability of success (being attacked by a bear) in one trial = 0.25 × 10^-6
x = number of successes (being attacked by a bear) in n trials = at least 1 attack
The probability of at least 1 attack in 20 years can be calculated as the complement of the probability of no attacks in 20 years, which is given by:
P(no attacks in 20 years) = (1 - p)^n
Substituting the values, we get:
P(no attacks in 20 years) = (1 - 0.25 × 10^-6)^20 ≈ 0.999995
Therefore, the probability of at least 1 attack in 20 years is approximately:
P(at least 1 attack in 20 years) = 1 - P(no attacks in 20 years) ≈ 1 - 0.999995 ≈ 0.000005
This means that the probability of being attacked by a bear at least once in 20 years of camping is very low, approximately 0.0005%. However, it is still important to take appropriate precautions while camping in bear country, such as storing food properly and carrying bear spray.
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The price-demand equation for gasoline is 0.2x+2p=60 where p is the price per gallon in dollars and x is the daily demand measured in millions of gallons.
a. What price should be charged if the demand is 40 million gallons?.
b. If the price increases by $0.5, by how much does the demand decrease?
a. To determine the price that should be charged if the demand is 40 million gallons, we need to substitute the given demand value into the price-demand equation and solve for p.
The price-demand equation is given as 0.2x + 2p = 60, where x represents the daily demand in millions of gallons and p represents the price per gallon in dollars.
Substituting x = 40 into the equation, we have:
0.2(40) + 2p = 60
8 + 2p = 60
2p = 60 - 8
2p = 52
p = 52/2
p = 26
Therefore, the price that should be charged if the demand is 40 million gallons is $26 per gallon.
b. To determine the decrease in demand resulting from a price increase of $0.5, we need to calculate the change in demand caused by the change in price.
The given price-demand equation is 0.2x + 2p = 60. Let's assume the initial price is p1 and the initial demand is x1. The new price is p2 = p1 + 0.5 (increase of $0.5), and we need to find the change in demand, Δx.
Substituting the initial price and demand into the equation, we have:
0.2x1 + 2p1 = 60
Now, substituting the new price and demand into the equation, we have:
0.2x2 + 2p2 = 60
To find the change in demand, we subtract the two equations:
(0.2x2 + 2p2) - (0.2x1 + 2p1) = 0
Simplifying the equation:
0.2x2 - 0.2x1 + 2p2 - 2p1 = 0
Since p2 = p1 + 0.5, we can substitute it in:
0.2x2 - 0.2x1 + 2(p1 + 0.5) - 2p1 = 0
0.2x2 - 0.2x1 + 2p1 + 1 - 2p1 = 0
0.2x2 - 0.2x1 + 1 = 0
Rearranging the equation:
0.2(x2 - x1) = -1
Dividing both sides by 0.2:
x2 - x1 = -1/0.2
x2 - x1 = -5
Therefore, the demand decreases by 5 million gallons when the price increases by $0.5.
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gretchen goes to buy a dozen donuts from a donut store that sells five varieties of donuts. one of the varieties of donuts sold is chocolate. how many ways are there to select the donuts if she must have exactly one chocolate donut in her selection?
if Gretchen must have exactly one chocolate donut in her selection, there are 330 ways to select 11 donuts from 4 varieties.
Ways of selecting on chocolate donut explainedNote, If Gretchen must have exactly one chocolate donut in her selection, then there are 11 remaining donuts to choose from, and she can choose any combination of the remaining four varieties of donuts.
We can use the combination formula to calculate the number of ways to choose 11 donuts from 4 varieties
C(11,4) = 11! / (4! * (11-4)!) = 330
Thus, there are 330 ways to select 11 donuts from 4 varieties.
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Suppose the supply and demand equations for printed T-shirts in a particular week are p=0. 7q+3 p=-1. 7q+10 Where p is the price in dollars and q is the quantity in hundreds. Find the equilibrium quantity
The equilibrium quantity is approximately 2.92 hundred T-shirts.
To find the equilibrium quantity, we need to set the supply and demand equations equal to each other and solve for q.
The supply equation is [tex]p = 0.7q + 3[/tex], where p is the price in dollars and q is the quantity in hundreds.
The demand equation is [tex]p = -1.7q + 10[/tex].
Setting them equal, we get [tex]0.7q + 3 = -1.7q + 10[/tex].
To solve for q, we can simplify the equation by adding 1.7q to both sides: [tex]2.4q + 3 = 10[/tex].
Then, subtracting 3 from both sides gives us [tex]2.4q = 7[/tex].
Finally, dividing both sides by 2.4 gives us [tex]q \approx 2.92[/tex].
Therefore, the equilibrium quantity is approximately 2.92 hundred T-shirts.
Please note that the actual quantity might not be exactly 2.92 hundred T-shirts due to rounding. Also, keep in mind that this is a hypothetical scenario and may not reflect real-world market dynamics.
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Evaluate yyye y 2 dv, where e is the solid hemisphere x 2 1 y 2 1 z2 < 9, y > 0.
The result of the triple integral is: [tex]I_E= [\frac{162}{5} ][/tex]
The solid E is the hemisphere of radius 3. It is the right part of the sphere
[tex]x^{2} +y^2+z^2=9[/tex] of radius 3 that corresponds to [tex]y\geq 0[/tex]
Here we slightly modify the spherical coordinates using the y axis as the azimuthal axis as this is more suitable for the given region. That is we interchange the roles of z and y in the standard spherical coordinate configuration. Now the angle [tex]\theta[/tex] is the polar angle on the xz plane measured from the positive x axis and [tex]\phi[/tex] is the azimuthal angle measured from the y axis.
Then the region can be parametrized as follows:
[tex]x=rcos\thetasin\phi\\\\y=rcos\phi\\\\z=rsin\theta\,sin\phi[/tex]
where the ranges of the variables are:
[tex]0\leq r\leq 3\\\\0\leq \theta\leq \pi \\\\0\leq \phi\leq \pi /2[/tex]
Calculate the triple integral. In the method of change of coordinates in triple integration we need the Jacobian of the transformation that is used to transform the volume element. We have,
[tex]J=r^2sin\phi \,\,\,\,\,[Jacobian \, of \,the\, transformation][/tex]
[tex]y^2=r^2cos^2\phi[/tex]
[tex]I_E=\int\int\int_E y^2dV[/tex]
[tex]I_E=\int_0^2^\pi \int^3_0\int_0^\\\pi /2[/tex][tex](r^2cos^2\phi)(r^2sin\phi)d\phi\, dr\, d\theta[/tex]
[tex]I_E=\int_0^2^\pi \int^3_0\int_0^\\\pi /2[/tex] [tex](r^4cos^2\phi sin\phi)d\phi\, dr\, d\theta[/tex]
Substitute [tex]u=cos \phi, du = -sin\phi \, du[/tex]
[tex]I_E=\int_0^2^\pi \int^3_0[-\frac{r^4}{3}cos^3\phi ]_0^\\\pi /2[/tex][tex]dr \, d\theta[/tex]
[tex]I_E=\int_0^2^\pi \int^3_0(\frac{r^4}{3} )dr \, d\theta[/tex]
[tex]I_E=\int_0^2^\pi [\frac{r^5}{15} ]^3_0 \, d\theta[/tex]
[tex]I_E=\int_0^2^\pi [\frac{3^5}{15} ] \, d\theta[/tex]
[tex]I_E= [\frac{81}{5}\theta ][/tex]
[tex]I_E= [\frac{81}{5}(2\theta) ]\\\\I_E= [\frac{162}{5} ][/tex]
The result of the triple integral is: [tex]I_E= [\frac{162}{5} ][/tex]
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Complete question is:
Evaluate [tex]\int\int_E\int y^2 \, dV[/tex] , where E is the solid hemisphere [tex]x^{2} +y^2+z^2=9[/tex], [tex]y\geq 0[/tex]
An engineer has designed a valve that will regulate water pressure on an automobile engine. The valve was tested on 120 engines and the mean pressure was 4.7lb/square inch. Assume the variance is known to be 0.81. If the valve was designed to produce a mean pressure of 4.9 lbs/square inch, is there sufficient evidence at the 0.02 level that the valve performs below the specifications? State the null and alternative hypotheses for the above scenario.
The engineer wants to test if there is sufficient evidence to support the claim that the valve performs below the specifications, which means they are interested in finding evidence to reject the null hypothesis in favor of the alternative hypothesis.
The null and alternative hypotheses for the scenario are as follows:
Null hypothesis (H0): The mean pressure produced by the valve is equal to or greater than the specified mean pressure of 4.9 lbs/square inch.
Alternative hypothesis (Ha): The mean pressure produced by the valve is below the specified mean pressure of 4.9 lbs/square inch.
Mathematically, it can be represented as:
H0: μ >= 4.9
Ha: μ < 4.9
Where μ represents the population mean pressure produced by the valve.
The engineer wants to test if there is sufficient evidence to support the claim that the valve performs below the specifications, which means they are interested in finding evidence to reject the null hypothesis in favor of the alternative hypothesis.
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f ′′ (t)−2f ′ (t)+2f(t)=0,f(π)=e π ,f ′ (π)=0 f(t)=
The solution to the differential equation that satisfies the initial conditions is: f(t) = -(1/4)e^π(1 + sqrt(2))*sin(sqrt(2)/2 *(t - π)) + (1/4)e^π(sqrt(2) - 1)*cos(sqrt(2)/2 *(t - π))
The given differential equation is:
f''(t) - 2f'(t) + 2f(t) = 0
We can write the characteristic equation as:
r^2 - 2r + 2 = 0
Solving this quadratic equation yields:
r = (2 ± sqrt(2)i)/2
The general solution to the differential equation is then:
f(t) = c1e^(r1t) + c2e^(r2t)
where r1 and r2 are the roots of the characteristic equation, and c1 and c2 are constants that we need to determine.
Since the roots of the characteristic equation are complex, we can express them in polar form as:
r1 = e^(ipi/4)
r2 = e^(-ipi/4)
Using Euler's formula, we can write these roots as:
r1 = (sqrt(2)/2 + isqrt(2)/2)
r2 = (sqrt(2)/2 - isqrt(2)/2)
Therefore, the general solution is:
f(t) = c1e^[(sqrt(2)/2 + isqrt(2)/2)t] + c2e^[(sqrt(2)/2 - i*sqrt(2)/2)*t]
To find the values of c1 and c2, we use the initial conditions f(π) = e^π and f'(π) = 0. First, we evaluate f(π):
f(π) = c1e^[(sqrt(2)/2 + isqrt(2)/2)π] + c2e^[(sqrt(2)/2 - isqrt(2)/2)π]
= c1(-1/2 + i/2) + c2(-1/2 - i/2)
Taking the real part of this equation and equating it to e^π, we get:
c1*(-1/2) + c2*(-1/2) = e^π / 2
Taking the imaginary part of the equation and equating it to zero (since f'(π) = 0), we get:
c1*(1/2) + c2*(-1/2) = 0
Solving these equations simultaneously, we get:
c1 = -(1/4)*e^π - (1/4)*sqrt(2)*e^π
c2 = (1/4)*sqrt(2)*e^π - (1/4)*e^π
Therefore, the solution to the differential equation that satisfies the initial conditions is:
f(t) = -(1/4)e^π(1 + sqrt(2))*sin(sqrt(2)/2 *(t - π)) + (1/4)e^π(sqrt(2) - 1)*cos(sqrt(2)/2 *(t - π))
Note that we have used Euler's formula to write the solution in terms of sines and cosines.
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Given that the current in a circuit is represented by the following equation, find the first time at which the current is a maximum. i=sin ^2
(4πt)+2sin(4πt)
The first time at which the current is a maximum is 0.125 seconds.
The equation that represents the current in a circuit is given by
i = sin²(4πt) + 2sin(4πt).
We need to find the first time at which the current is a maximum.
We can re-write the given equation by substituting
sin(4πt) = x.
Then, i = sin²(4πt) + 2sin(4πt) = x² + 2x
Differentiating both sides with respect to time, we get
di/dt = (d/dt)(x² + 2x) = 2x dx/dt + 2 dx/dt
where x = sin(4πt)
Thus, di/dt = 2sin(4πt) (4π cos(4πt) + 1)
Now, for current to be maximum, di/dt = 0
Therefore, 2sin(4πt) (4π cos(4πt) + 1) = 0or sin(4πt) (4π cos(4πt) + 1) = 0
Either sin(4πt) = 0 or 4π cos(4πt) + 1 = 0
We know that sin(4πt) = 0 at t = 0, 0.25, 0.5, 0.75, 1.0, 1.25 seconds.
However, sin(4πt) = 0 gives minimum current, not maximum.
Hence, we consider the second equation.4π cos(4πt) + 1 = 0cos(4πt) = -1/4π
At the first instance of cos(4πt) = -1/4π, i.e. when t = 0.125 seconds, the current will be maximum.
Hence, the first time at which the current is a maximum is 0.125 seconds.
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A student’s first 3 grades are 70, 82, and 94. What grade must she make on the 4th texts to have an average of all 4 tests of 80? Identify the unknown, set up an equation and use Algebra to solve. Show all 4 steps. (only half credit possible if you do not set up an algebraic equation to solve)
The student must score 74 on the fourth test to have an average of 80 for all four tests, The equation can be formed by considering the average of the four tests,
To find the grade the student must make on the fourth test to achieve an average of 80 for all four tests, we can set up an algebraic equation. Let the unknown grade on the fourth test be represented by "x."
The equation can be formed by considering the average of the four tests, which is obtained by summing up all the grades and dividing by 4. By rearranging the equation and solving for "x," we can determine that the student needs to score 84 on the fourth test to achieve an average of 80 for all four tests.
Let's denote the unknown grade on the fourth test as "x." The average of all four tests can be calculated by summing up the grades and dividing by the total number of tests, which is 4.
In this case, the sum of the first three grades is 70 + 82 + 94 = 246. So, the equation representing the average is (70 + 82 + 94 + x) / 4 = 80.
To solve this equation, we can begin by multiplying both sides of the equation by 4 to eliminate the fraction: 70 + 82 + 94 + x = 320. Next, we can simplify the equation by adding up the known grades: 246 + x = 320.
To isolate "x," we can subtract 246 from both sides of the equation: x = 320 - 246. Simplifying further, we have x = 74.
Therefore, the student must score 74 on the fourth test to have an average of 80 for all four tests.
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Evaluating an algebraic expression: Whole nu Evaluate the expression when a=4 and c=2. (4c+a^(2))/(c)
The expression (4c+a^(2))/(c) when a=4 and c=2, we substitute the given values for a and c into the expression and simplify it using the order of operations.
Evaluate the expression (4c + a^2)/c when a = 4 and c = 2, we substitute the given values into the expression. First, we calculate the value of a^2: a^2 = 4^2 = 16. Then, we substitute the values of a^2, c, and 4c into the expression: (4c + a^2)/c = (4 * 2 + 16)/2 = (8 + 16)/2 = 24/2 = 12. Therefore, when a = 4 and c = 2, the expression (4c + a^2)/c evaluates to 12.
First, substitute a=4 and c=2 into the expression:
(4(2)+4^(2))/(2)
Next, simplify using the order of operations:
(8+16)/2
= 24/2
= 12
Therefore, the value of the expression (4c+a^(2))/(c) when a=4 and c=2 is 12.
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Find the syact solutions (in racians) to the equations in the given interval. Note - No thig identities are needed, And there are only two arawiers if each problem, enter single answers in each field. Warning: fio credit will be give for answers using inverse trig functions, degrees, or cafculator approximatians: (a) cos(θ)(cos(θ)−4)=0 for 0≤θ<2π =________ (b) (tan(x)−1) 2
=0 for 0⩽x⩽2x___________
(a) The solutions to the equation cos(θ)(cos(θ) - 4) = 0 in the interval 0 ≤ θ < 2π are θ = π/2 and θ = 3π/2. (b) The solution to the equation (tan(x) - 1)² = 0 in the interval 0 ≤ x ≤ 2π is x = π/4.
(a) The equation cos(θ)(cos(θ) - 4) = 0 can be rewritten as cos²(θ) - 4cos(θ) = 0. Factoring out cos(θ), we have cos(θ)(cos(θ) - 4) = 0.
Setting each factor equal to zero:
cos(θ) = 0 or cos(θ) - 4 = 0.
For the first factor, cos(θ) = 0, the solutions in the interval 0 ≤ θ < 2π are θ = π/2 and θ = 3π/2.
For the second factor, cos(θ) - 4 = 0, we have cos(θ) = 4, which has no real solutions since the range of cosine function is -1 to 1.
(b) The equation (tan(x) - 1)² = 0 can be expanded as tan²(x) - 2tan(x) + 1 = 0.
Setting each term equal to zero:
tan²(x) - 2tan(x) + 1 = 0.
Factoring the equation, we have (tan(x) - 1)(tan(x) - 1) = 0.
Setting each factor equal to zero:
tan(x) - 1 = 0.
Solving for x, we have x = π/4.
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In Python 3. The Fibonacci sequence is defined as follows: f 1
=1
f 2
=1
f n
=f n−1
+f n−2
for n>2
The first few numbers of the sequence are: 1,1,2,3,5,8… A Fibonacci number is any number found in this sequence. Note that this definition does not consider 0 to be a Fibonacci number. Given a list of numbers, determine if each number is the sum of two Fibonacci numbers. Example Given an input of [2,5,17], the function is expected to return This is because 1+1=2,2+3=5 but there are no two Fibonacci numbers that sum to 17 . - [execution time limit] 4 seconds (py3) - [input] array.integer64 a A list of numbers which we want to query. The length is guaranteed to be less than 5000. 1≤a i
≤10 18
- [output] array.boolean List of booleans, b, where each element b i
corresponds to the answer to query a i
.
Here is the Python code for the given problem statement:
```
def is_fib(n):
if n == 0:
return False
a, b = 1, 1
while b < n:
a, b = b, a + b
return b == n
def sum_fib(n):
a, b = 1, 1
while a <= n:
if is_fib(n - a):
return True
a, b = b, a + b
return False
def fibonacci_sum(a):
return [sum_fib(n) for n in a]```
The function is_fib checks if a given number n is a Fibonacci number or not. The function sum_fib checks if a given number n is the sum of two Fibonacci numbers or not.
The function fibonacci_sum returns a list of booleans corresponding to whether each number in the input list is the sum of two Fibonacci numbers or not.
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