The function [tex]\(f(x)=\frac{x^{2}}{x-2}\)[/tex] has increasing intervals from negative infinity to 2 and from 2 to positive infinity.
To find the intervals where the function f(x) is increasing, we need to determine where its derivative is positive. Let's start by finding the derivative of f(x): [tex]\[f'(x) = \frac{d}{dx}\left(\frac{x^{2}}{x-2}\right)\][/tex]
Using the quotient rule, we can differentiate the function:
[tex]\[f'(x) = \frac{(x-2)(2x) - (x^2)(1)}{(x-2)^2}\][/tex]
Simplifying this expression gives us:
[tex]\[f'(x) = \frac{2x^2 - 4x - x^2}{(x-2)^2}\][/tex]
[tex]\[f'(x) = \frac{x^2 - 4x}{(x-2)^2}\][/tex]
[tex]\[f'(x) = \frac{x(x-4)}{(x-2)^2}\][/tex]
To determine where the derivative is positive, we consider the sign of f'(x). The function f'(x) will be positive when both x(x-4) and (x-2)² have the same sign. Analyzing the sign of each factor, we can determine the intervals:
x(x-4) is positive when x < 0 or x > 4.
(x-2)^2 is positive when x < 2 or x > 2.
Since both factors have the same sign for x < 0 and x > 4, and x < 2 and x > 2, we can conclude that the function f(x) is increasing on the intervals from negative infinity to 2 and from 2 to positive infinity.
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Solve the system. x1−6x32x1+2x2+3x3x2+4x3=22=11=−6 Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. The unique solution of the system is । (Type integers or simplified fractions.) B. The system has infinitely many solutions. C. The system has no solution.
The unique solution for the system x1−6x32x1+2x2+3x3x2+4x3=22=11=−6 is given system of equations is x1 = -3, x2 = 7, and x3 = 6. Thus, Option A is the answer.
We can write the system of linear equations as:| 1 - 6 0 | | x1 | | 2 || 2 2 3 | x | x2 | = |11| | 0 1 4 | | x3 | |-6 |
Let A = | 1 - 6 0 || 2 2 3 || 0 1 4 | and,
B = | 2 ||11| |-6 |.
Then, the system of equations can be written as AX = B.
Now, we need to find the value of X.
As AX = B,
X = A^(-1)B.
Thus, we can find the value of X by multiplying the inverse of A and B.
Let's find the inverse of A:| 1 - 6 0 | | 2 0 3 | |-18 6 2 || 2 2 3 | - | 0 1 0 | = | -3 1 -1 || 0 1 4 | | 0 -4 2 | | 2 -1 1 |
Thus, A^(-1) = | -3 1 -1 || 2 -1 1 || 2 0 3 |
We can multiply A^(-1) and B to get the value of X:
| -3 1 -1 | | 2 | | -3 | | 2 -1 1 | |11| | 7 |X = | 2 -1 1 | * |-6| = |-3 || 2 0 3 | |-6| | 6 |
Thus, the solution of the given system of equations is x1 = -3, x2 = 7, and x3 = 6.
Therefore, the unique solution of the system is A.
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Find a polynomial function that has the given zeros. (There are many correct answers.) \[ 4,-5,5,0 \] \[ f(x)= \]
A polynomial function with zeros 4, -5, 5, and 0 is f(x) = 0.
To find a polynomial function with zeros 4, -5, 5, and 0, we need to start with a factored form of the polynomial. The factored form of a polynomial with these zeros is:
f(x) = a(x - 4)(x + 5)(x - 5)x
where a is a constant coefficient.
To find the value of a, we can use any of the known points of the polynomial. Since the polynomial has a zero at x = 0, we can substitute x = 0 into the factored form and solve for a:
f(0) = a(0 - 4)(0 + 5)(0 - 5)(0) = 0
Simplifying this equation, we get:
0 = -500a
Therefore, a = 0.
Substituting this into the factored form, we get:
f(x) = 0(x - 4)(x + 5)(x - 5)x = 0
Therefore, a polynomial function with zeros 4, -5, 5, and 0 is f(x) = 0.
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Find and classify the critical points of z=(x 2
−2x)(y 2
−7y) Local maximums: Local minimums: Saddle points: For each classification, enter a list of ordered pairs (x,y) where the max/min/saddle occurs. If there are no points for a classification, enter DNE.
A critical point is a point at which the first derivative is zero or the second derivative test is inconclusive.
A critical point is a stationary point at which a function's derivative is zero. When finding the critical points of the function z = (x2−2x)(y2−7y), we'll use the second derivative test to classify them as local maxima, local minima, or saddle points. To begin, we'll find the partial derivatives of the function z with respect to x and y, respectively, and set them equal to zero to find the critical points.∂z/∂x = 2(x−1)(y2−7y)∂z/∂y = 2(y−3)(x2−2x)
Setting the above partial derivatives to zero, we have:2(x−1)(y2−7y) = 02(y−3)(x2−2x) = 0
Therefore, we get x = 1 or y = 0 or y = 7 or x = 0 or x = 2 or y = 3.
After finding the values of x and y, we must find the second partial derivatives of z with respect to x and y, respectively.∂2z/∂x2 = 2(y2−7y)∂2z/∂y2 = 2(x2−2x)∂2z/∂x∂y = 4xy−14x+2y2−42y
If the second partial derivative test is negative, the point is a maximum. If it's positive, the point is a minimum. If it's zero, the test is inconclusive. And if both partial derivatives are zero, the test is inconclusive. Therefore, we use the second derivative test to classify the critical points into local minima, local maxima, and saddle points.
∂2z/∂x2 = 2(y2−7y)At (1, 0), ∂2z/∂x2 = 0, which is inconclusive.
∂2z/∂x2 = 2(y2−7y)At (1, 7), ∂2z/∂x2 = 0, which is inconclusive.∂2z/∂x2 = 2(y2−7y)At (0, 3), ∂2z/∂x2 = −42, which is negative and therefore a local maximum.
∂2z/∂x2 = 2(y2−7y)At (2, 3), ∂2z/∂x2 = 42, which is positive and therefore a local minimum.
∂2z/∂y2 = 2(x2−2x)At (1, 0), ∂2z/∂y2 = −2, which is a saddle point.
∂2z/∂y2 = 2(x2−2x)At (1, 7), ∂2z/∂y2 = 2, which is a saddle point.
∂2z/∂y2 = 2(x2−2x)
At (0, 3), ∂2z/∂y2 = 0, which is inconclusive.∂2z/∂y2 = 2(x2−2x)At (2, 3), ∂2z/∂y2 = 0, which is inconclusive.
∂2z/∂x∂y = 4xy−14x+2y2−42yAt (1, 0), ∂2z/∂x∂y = 0, which is inconclusive.
∂2z/∂x∂y = 4xy−14x+2y2−42yAt (1, 7), ∂2z/∂x∂y = 0, which is inconclusive.
∂2z/∂x∂y = 4xy−14x+2y2−42yAt (0, 3), ∂2z/∂x∂y = −14, which is negative and therefore a saddle point.
∂2z/∂x∂y = 4xy−14x+2y2−42yAt (2, 3), ∂2z/∂x∂y = 14, which is positive and therefore a saddle point. Therefore, we obtain the following classification of critical points:Local maximums: (0, 3)Local minimums: (2, 3)
Saddle points: (1, 0), (1, 7), (0, 3), (2, 3)
Thus, using the second derivative test, we can classify the critical points as local maxima, local minima, or saddle points. At the local maximum and local minimum points, the function's partial derivatives with respect to x and y are both zero. At the saddle points, the function's partial derivatives with respect to x and y are not equal to zero. Furthermore, the second partial derivative test, which evaluates the signs of the second-order partial derivatives of the function, is used to classify the critical points as local maxima, local minima, or saddle points. Critical points of the given function are (0, 3), (2, 3), (1, 0), (1, 7).These points have been classified as local maximum, local minimum and saddle points.The local maximum point is (0, 3)The local minimum point is (2, 3)The saddle points are (1, 0), (1, 7), (0, 3), (2, 3).
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Question 4 (a) Prove by mathematical induction that \( n^{3}+5 n \) is divisible by 6 for all \( n=1,2,3, \ldots \) [9 marks]
We will prove by mathematical induction that [tex]n^3 +5n[/tex] is divisible by 6 for all positive integers [tex]n[/tex].
To prove the divisibility of [tex]n^3 +5n[/tex] by 6 for all positive integers [tex]n[/tex], we will use mathematical induction.
Base Case:
For [tex]n=1[/tex], we have [tex]1^3 + 5*1=6[/tex], which is divisible by 6.
Inductive Hypothesis:
Assume that for some positive integer [tex]k, k^3+5k[/tex] is divisible by 6.
Inductive Step:
We need to show that if the hypothesis holds for k, it also holds for k+1.
Consider,
[tex](k+1)^3+5(k+1)=k ^3+3k^2+3k+1+5k+5[/tex]
By the inductive hypothesis, we know that 3+5k is divisible by 6.
Additionally, [tex]3k^2+3k[/tex] is divisible by 6 because it can be factored as 3k(k+1), where either k or k+1 is even.
Hence, [tex](k+1)^3 +5(k+1)[/tex] is also divisible by 6.
Since the base case holds, and the inductive step shows that if the hypothesis holds for k, it also holds for k+1, we can conclude by mathematical induction that [tex]n^3 + 5n[/tex] is divisible by 6 for all positive integers n.
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The water-supply manager for dallas needs to supply the city with at least 19 million gallons of potable water per day. the supply may be drawn from the local reservoir or from a pipeline to an adjacent town. the local reservoir has a maximum daily yield of 20 million gallons of potable water, and the pipeline has a maximum daily yield of 13 million gallons. by contract, the pipeline is required to supply a minimum of 7 million gallons per day. if the cost for 1 million gallons of reservoir water is $290 and the cost for 1 million gallons of pipeline water is $365, how much water should the manager get from each source to minimize daily water costs for the city? what is the minimum daily water cost?
So, the manager should get all the required water from the local reservoir, resulting in a minimum daily water cost of $5510.
To minimize the daily water costs for the city, the water-supply manager needs to determine how much water to get from each source while meeting the minimum requirement of 19 million gallons per day. Let's denote the amount of water drawn from the local reservoir as R (in million gallons) and the amount of water drawn from the pipeline as P (in million gallons).
Given the constraints:
R ≤ 20 (maximum daily yield of the reservoir)
P ≥ 7 (minimum daily yield of the pipeline)
R + P ≥ 19 (minimum requirement of 19 million gallons)
We need to find the values of R and P that satisfy these constraints while minimizing the daily water costs.
Let's calculate the costs for each source:
Cost of 1 million gallons of reservoir water = $290
Cost of 1 million gallons of pipeline water = $365
The total daily cost can be expressed as:
Total Cost = (Cost of reservoir water per million gallons) * R + (Cost of pipeline water per million gallons) * P
To minimize the total cost, we can use linear programming techniques or analyze the possible combinations. In this case, since the costs per million gallons are provided, we can directly compare the costs and evaluate the options.
Let's consider a few scenarios:
If all the water (19 million gallons) is drawn from the reservoir:
Total Cost = (Cost of reservoir water per million gallons) * 19 = $290 * 19
If all the water (19 million gallons) is drawn from the pipeline:
Total Cost = (Cost of pipeline water per million gallons) * 19 = $365 * 19
If some water is drawn from the reservoir and the remaining from the pipeline: Since the minimum requirement is 19 million gallons, the pipeline must supply at least 19 - 20 = -1 million gallons, which is not possible. Thus, this scenario is not valid. Therefore, to minimize the daily water costs, the manager should draw all 19 million gallons of water from the local reservoir. The minimum daily water cost would be:
Minimum Daily Water Cost = (Cost of reservoir water per million gallons) * 19 = $290 * 19 = $5510.
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Set up the integral of \( f(r, \theta, z)=r_{z} \) oven the region bounded above by the sphere \( r^{2}+z^{2}=2 \) and bounded below by the cone \( z=r \)
We have to set up the integral of \(f(r, \theta, z) = r_z\) over the region bounded above by the sphere \(r^2 + z^2 = 2\) and bounded below by the cone \(z = r\).The given region can be shown graphically as:
The intersection curve of the cone and sphere is a circle at \(z = r = 1\). The sphere completely encloses the cone, thus we can set the limits of integration from the cone to the sphere, i.e., from \(r\) to \(\sqrt{2 - z^2}\), and from \(0\) to \(\pi/4\) in the \(\theta\) direction. And from \(0\) to \(1\) in the \(z\) direction.
So, the integral to evaluate is given by:\iiint f(r, \theta, z) dV = \int_{0}^{\pi/4} \int_{0}^{2\pi} \int_{0}^{1} \frac{\partial r}{\partial z} r \, dr \, d\theta \, dz= \int_{0}^{\pi/4} \int_{0}^{2\pi} \int_{0}^{1} \frac{z}{\sqrt{2 - z^2}} r \, dr \, d\theta \, dz= 2\pi \int_{0}^{1} \int_{z}^{\sqrt{2 - z^2}} \frac{z}{\sqrt{2 - z^2}} r \, dr \, dz= \pi \int_{0}^{1} \left[ \sqrt{2 - z^2} - z^2 \ln\left(\sqrt{2 - z^2} + \sqrt{z^2}\right) \right] dz= \pi \left[ \frac{\pi}{4} - \frac{1}{3}\sqrt{3} \right]the integral of \(f(r, \theta, z) = r_z\) over the given region is \(\pi \left[ \frac{\pi}{4} - \frac{1}{3}\sqrt{3} \right]\).
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State whether the following statement is true or false. The two lines 5x+y=5 and 10x+2y=0 are parallel. Choose the correct answer below. True False
The correct answer that they are parallel or not is: True.
To determine if two lines are parallel, we need to compare their slopes. If the slopes of two lines are equal, then the lines are parallel.
If the slopes are different, the lines are not parallel.
Let's analyze the given lines:
Line 1: 5x + y = 5
Line 2: 10x + 2y = 0
To compare the slopes, we need to rewrite the equations in slope-intercept form (y = mx + b), where "m" represents the slope:
Line 1:
5x + y = 5
y = -5x + 5
Line 2:
10x + 2y = 0
2y = -10x
y = -5x
By comparing the slopes, we can see that the slopes of both lines are equal to -5. Since the slopes are the same, we can conclude that the lines are indeed parallel.
Therefore, the correct answer that they are parallel or not: True.
It's important to note that parallel lines have the same slope but may have different y-intercepts. In this case, both lines have a slope of -5, indicating that they are parallel.
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Find the slope of the line if it exists.
Answer:
m = -4/3
Step-by-step explanation:
Slope = rise/run or (y2 - y1) / (x2 - x1)
Pick 2 points (-2,2) (1,-2)
We see the y decrease by 4 and the x increase by 3, so the slope is
m = -4/3
Abody moves on a coordinate line such that it has a position s =f(t)=t 2 −3t+2 on the interval 0≤t≤9, with sin meters and t in seconds. a. Find the body's displacement and average velocity for the given time interval. b. Find the body's speed and acceleration at the endpoints of the interval. c. When, if ever, during the interval does the body change direction?
The body's displacement on the interval 0 ≤ t ≤ 9 is 56 meters, and the average velocity is 6.22 m/s. The body's speed at t = 0 is 3 m/s, and at t = 9 it is 15 m/s. The acceleration at both endpoints is 2 m/s². The body changes direction at t = 3/2 seconds during the interval 0 ≤ t ≤ 9.
a. To determine the body's displacement on the interval 0 ≤ t ≤ 9, we need to evaluate f(9) - f(0):
Displacement = f(9) - f(0) = (9^2 - 3*9 + 2) - (0^2 - 3*0 + 2) = (81 - 27 + 2) - (0 - 0 + 2) = 56 meters
To determine the average velocity, we divide the displacement by the time interval:
Average velocity = Displacement / Time interval = 56 meters / 9 seconds = 6.22 m/s (rounded to two decimal places)
b. To ]determinine the body's speed at the endpoints of the interval, we calculate the magnitude of the velocity. The velocity is the derivative of the position function:
v(t) = f'(t) = 2t - 3
Speed at t = 0: |v(0)| = |2(0) - 3| = 3 m/s
Speed at t = 9: |v(9)| = |2(9) - 3| = 15 m/s
To determine the acceleration at the endpoints, we take the derivative of the velocity function:
a(t) = v'(t) = 2
Acceleration at t = 0: a(0) = 2 m/s²
Acceleration at t = 9: a(9) = 2 m/s²
c. The body changes direction whenever the velocity changes sign. In this case, we need to find when v(t) = 0:
2t - 3 = 0
2t = 3
t = 3/2
Therefore, the body changes direction at t = 3/2 seconds during the interval 0 ≤ t ≤ 9.
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Find the general solution to the following differential equations:
16y''-8y'+y=0
y"+y'-2y=0
y"+y'-2y = x^2
The general solution of the given differential equations are:
y = c₁e^(x/4) + c₂xe^(x/4) (for 16y''-8y'+y=0)
y = c₁e^x + c₂e^(-2x) (for y"+y'-2y=0)
y = c₁e^x + c₂e^(-2x) + (1/2)x
(for y"+y'-2y=x²)
Given differential equations are:
16y''-8y'+y=0
y"+y'-2y=0
y"+y'-2y = x²
To find the general solution to the given differential equations, we will solve these equations one by one.
(i) 16y'' - 8y' + y = 0
The characteristic equation is:
16m² - 8m + 1 = 0
Solving this quadratic equation, we get m = 1/4, 1/4
Hence, the general solution of the given differential equation is:
y = c₁e^(x/4) + c₂xe^(x/4)..................................................(1)
(ii) y" + y' - 2y = 0
The characteristic equation is:
m² + m - 2 = 0
Solving this quadratic equation, we get m = 1, -2
Hence, the general solution of the given differential equation is:
y = c₁e^x + c₂e^(-2x)..................................................(2)
(iii) y" + y' - 2y = x²
The characteristic equation is:
m² + m - 2 = 0
Solving this quadratic equation, we get m = 1, -2.
The complementary function (CF) of this differential equation is:
y = c₁e^x + c₂e^(-2x)..................................................(3)
Now, we will find the particular integral (PI). Let's assume that the PI of the differential equation is of the form:
y = Ax² + Bx + C
Substituting the value of y in the given differential equation, we get:
2A - 4A + 2Ax² + 4Ax - 2Ax² = x²
Equating the coefficients of x², x, and the constant terms on both sides, we get:
2A - 2A = 1,
4A - 4A = 0, and
2A = 0
Solving these equations, we get
A = 1/2,
B = 0, and
C = 0
Hence, the particular integral of the given differential equation is:
y = (1/2)x²..................................................(4)
The general solution of the given differential equation is the sum of CF and PI.
Hence, the general solution is:
y = c₁e^x + c₂e^(-2x) + (1/2)x²..................................................(5)
Conclusion: Therefore, the general solution of the given differential equations are:
y = c₁e^(x/4) + c₂xe^(x/4) (for 16y''-8y'+y=0)
y = c₁e^x + c₂e^(-2x) (for y"+y'-2y=0)
y = c₁e^x + c₂e^(-2x) + (1/2)x
(for y"+y'-2y=x²)
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The particular solution is: y = -1/2 x². The general solution is: y = c1 e^(-2x) + c2 e^(x) - 1/2 x²
The general solution of the given differential equations are:
Given differential equation: 16y'' - 8y' + y = 0
The auxiliary equation is: 16m² - 8m + 1 = 0
On solving the above quadratic equation, we get:
m = 1/4, 1/4
∴ General solution of the given differential equation is:
y = c1 e^(x/4) + c2 x e^(x/4)
Given differential equation: y" + y' - 2y = 0
The auxiliary equation is: m² + m - 2 = 0
On solving the above quadratic equation, we get:
m = -2, 1
∴ General solution of the given differential equation is:
y = c1 e^(-2x) + c2 e^(x)
Given differential equation: y" + y' - 2y = x²
The auxiliary equation is: m² + m - 2 = 0
On solving the above quadratic equation, we get:m = -2, 1
∴ The complementary solution is:y = c1 e^(-2x) + c2 e^(x)
Now we have to find the particular solution, let us assume the particular solution of the given differential equation:
y = ax² + bx + c
We will use the method of undetermined coefficients.
Substituting y in the differential equation:y" + y' - 2y = x²a(2) + 2a + b - 2ax² - 2bx - 2c = x²
Comparing the coefficients of x² on both sides, we get:-2a = 1
∴ a = -1/2
Comparing the coefficients of x on both sides, we get:-2b = 0 ∴ b = 0
Comparing the constant terms on both sides, we get:2c = 0 ∴ c = 0
Thus, the particular solution is: y = -1/2 x²
Now, the general solution is: y = c1 e^(-2x) + c2 e^(x) - 1/2 x²
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The function s=f(t) gives the position of a body moving on a coordinate line, with s in meters and t in seconds. Find the body's speed and acceleration at the end of the time interval. s=−t 3
+4t 2
−4t,0≤t≤4 A. 20 m/sec,−4 m/sec 2
B. −20 m/sec ,
−16 m/sec 2
C. 4 m/sec,0 m/sec 2
D. 20 m/sec,−16 m/sec 2
The correct option is B. −20 m/sec, −16 m/sec^2, the speed of the body is the rate of change of its position,
which is given by the derivative of s with respect to t. The acceleration of the body is the rate of change of its speed, which is given by the second derivative of s with respect to t.
In this case, the velocity is given by:
v(t) = s'(t) = −3t^2 + 8t - 4
and the acceleration is given by: a(t) = v'(t) = −6t + 8
At the end of the time interval, t = 4, the velocity is:
v(4) = −3(4)^2 + 8(4) - 4 = −20 m/sec
and the acceleration is: a(4) = −6(4) + 8 = −16 m/sec^2
Therefore, the body's speed and acceleration at the end of the time interval are −20 m/sec and −16 m/sec^2, respectively.
The velocity function is a quadratic function, which means that it is a parabola. The parabola opens downward, which means that the velocity is decreasing. The acceleration function is a linear function, which means that it is a line.
The line has a negative slope, which means that the acceleration is negative. This means that the body is slowing down and eventually coming to a stop.
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An object is tossed vertically upward from ground level. Its height s(t), in feet, at time t seconds is given by the position function s=−16t 2
+144t. n how many seconds does the object return to the point from which it was thrown? sec
The object returns to the point from which it was thrown in 9 seconds.
To determine the time at which the object returns to the point from which it was thrown, we set the height function s(t) equal to zero, since the object would be at ground level at that point. The height function is given by s(t) = -16t² + 144t.
Setting s(t) = 0, we have:
-16t²+ 144t = 0
Factoring out -16t, we get:
-16t(t - 9) = 0
This equation is satisfied when either -16t = 0 or t - 9 = 0. Solving these equations, we find that t = 0 or t = 9.
However, since the object is tossed vertically upward, we are only interested in the positive time when it returns to the starting point. Therefore, the object returns to the point from which it was thrown in 9 seconds.
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A furniture manufacturer makes chairs and sets price according to the following equation, where p is the price and q is the quantity produced. p(q)=1600−8q Express, using functional notation, the set price when the manufacturer produces 50 chairs? p( What is the value returned from that function p ? A furniture manufacturer makes chairs and sets price according to the following equation, where p is the price and q is the quantity produced. p(q)=1600−8q Express, using functional notation, how many chairs should be produced to sell them at $ 1,000 each? p(75)p(1000)=75751000p(q)=75∘p(q)=1000 What is the value returned from that function (what is q )?
When the furniture manufacturer produces 50 chairs, the set price is $1200. To sell the chairs at $1000 each, the manufacturer should produce 75 chairs.
Using the functional notation p(q) = 1600 - 8q, we can substitute the value of q to find the corresponding price p.
a) For q = 50, we have:
p(50) = 1600 - 8(50)
p(50) = 1600 - 400
p(50) = 1200
Therefore, when the manufacturer produces 50 chairs, the set price is $1200.
b) To find the number of chairs that should be produced to sell them at $1000 each, we can set the equation p(q) = 1000 and solve for q.
p(q) = 1600 - 8q
1000 = 1600 - 8q
8q = 600
q = 600/8
q = 75
Hence, to sell the chairs at $1000 each, the manufacturer should produce 75 chairs.
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need help ive never done this before
For the following function find \( f(x+h) \) and \( f(x)+f(h) \). \[ f(x)=x^{2}-1 \] \( f(x+h)= \) (Simplify your answer.)
f(x+h) = (x+h)^2 - 1 = x^2 + 2hx + h^2 - 1, f(x+h) can be used to find the value of f(x) when x is increased by h.
To find f(x+h), we can substitute x+h into the function f(x) = x^2-1. This gives us f(x+h) = (x+h)^2 - 1
We can expand the square to get:
f(x+h) = x^2 + 2hx + h^2 - 1
Here is a more detailed explanation of how to find f(x+h):
Substitute x+h into the function f(x) = x^2-1. Expand the square. Simplify the expression.f(x+h) can be used to find the value of f(x) when x is increased by h. For example, if x = 2 and h = 1, then f(x+h) = f(3) = 9.
f(x)+f(h):
f(x)+f(h) = x^2-1 + h^2-1 = x^2+h^2-2
Here is a more detailed explanation of how to find f(x)+f(h):
Add f(x) and f(h).Simplify the expression.f(x)+f(h) can be used to find the sum of the values of f(x) and f(h). For example, if x = 2 and h = 1, then f(x)+f(h) = f(2)+f(1) = 5.
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A triangle is rightangled triangle if ad = 12 bd = dc then find the length of bd and dc
The length of bd (and dc) is approximately 8.49 units.
To find the length of bd and dc in a right-angled triangle with ad = 12, we can use the Pythagorean theorem. In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Let's label the sides of the triangle as follows:
- ad is the hypotenuse
- bd is one of the legs
- dc is the other leg
Using the Pythagorean theorem we have the equation:
(ad)² = (bd)² + (dc)²
Given that ad = 12, we can substitute it into the equation:
(12)² = (bd)² + (dc)²
Simplifying further:
144 = (bd)² + (dc)²
Since bd = dc (as mentioned in the question), we can substitute bd for dc:
144 = (bd)² + (bd)²
Combining like terms:
144 = 2(bd)²
Dividing both sides by 2:
72 = (bd)²
Taking the square root of both sides:
bd = √72
Simplifying:
bd ≈ 8.49
Therefore, the length of bd (and dc) is approximately 8.49 units.
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Let F=⟨0, z
x
,e −xyz
⟩ and let S be the portion of the paraboloid z=2−x 2
−y 2
,z≥−2, oriented upward. Use Stokes' Theorem to evaluate
Stokes' Theorem states that the line integral of a vector field F around a simple closed curve C is equal to the surface integral of the curl of F over the surface S bounded by C. In other words:
∮C F · dr = ∬S curl(F) · dS
In this case, the surface S is the portion of the paraboloid z = 2 - x^2 - y^2 for z ≥ -2, oriented upward. The boundary curve C of this surface is the circle x^2 + y^2 = 4 in the plane z = -2.
The curl of a vector field F = ⟨P, Q, R⟩ is given by:
curl(F) = ⟨Ry - Qz, Pz - Rx, Qx - Py⟩
For the vector field F = ⟨0, z/x, e^(-xyz)⟩, we have:
P = 0
Q = z/x
R = e^(-xyz)
Taking the partial derivatives of P, Q, and R with respect to x, y, and z, we get:
Px = 0
Py = 0
Pz = 0
Qx = -z/x^2
Qy = 0
Qz = 1/x
Rx = -yze^(-xyz)
Ry = -xze^(-xyz)
Rz = -xye^(-xyz)
Substituting these partial derivatives into the formula for curl(F), we get:
curl(F) = ⟨Ry - Qz, Pz - Rx, Qx - Py⟩
= ⟨-xze^(-xyz) - 1/x, 0 - (-yze^(-xyz)), -z/x^2 - 0⟩
= ⟨-xze^(-xyz) - 1/x, yze^(-xyz), -z/x^2⟩
To evaluate the surface integral of curl(F) over S using Stokes' Theorem, we need to parameterize the boundary curve C. Since C is the circle x^2 + y^2 = 4 in the plane z = -2, we can parameterize it as follows:
r(t) = ⟨2cos(t), 2sin(t), -2⟩ for 0 ≤ t ≤ 2π
The line integral of F around C is then given by:
∮C F · dr
= ∫(from t=0 to 2π) F(r(t)) · r'(t) dt
= ∫(from t=0 to 2π) ⟨0, (-2)/(2cos(t)), e^(4cos(t)sin(t))⟩ · ⟨-2sin(t), 2cos(t), 0⟩ dt
= ∫(from t=0 to 2π) [0*(-2sin(t)) + ((-2)/(2cos(t)))*(2cos(t)) + e^(4cos(t)sin(t))*0] dt
= ∫(from t=0 to 2π) (-4 + 0 + 0) dt
= ∫(from t=0 to 2π) (-4) dt
= [-4t] (from t=0 to 2π)
= **-8π**
Therefore, by Stokes' Theorem, the surface integral of curl(F) over S is equal to **-8π**.
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Use the given function and the given interval to complete parts a and b. f(x)=2x 3−33x 2 +144x on [2,9] a. Determine the absolute extreme values of f on the given interval when they exist. b. Use a graphing utility to confirm your conclusions. a. What is/are the absolute maximum/maxima of fon the given interval? Select the correct choice below and, if necessary, fill in the answer box to complete your choice. A. The absolute maximum/maxima is/are at x= (Use a comma to separate answers as needed. Type exact answers, using radicals as needed.) B. There is no absolute maximum of f on the given interval.
The absolute maximum of the function \(f(x) = 2x^3 - 33x^2 + 144x\) on the interval \([2, 9]\) is 297.
a. The absolute maximum of \(f\) on the given interval is at \(x = 9\).
b. Graphing utility can be used to confirm this conclusion by plotting the function \(f(x)\) over the interval \([2, 9]\) and observing the highest point on the graph.
To determine the absolute extreme values of the function \(f(x) = 2x^3 - 33x^2 + 144x\) on the interval \([2, 9]\), we can follow these steps:
1. Find the critical points of the function within the given interval by finding where the derivative equals zero or is undefined.
2. Evaluate the function at the critical points and the endpoints of the interval.
3. Identify the highest and lowest values among the critical points and the endpoints to determine the absolute maximum and minimum.
Let's begin with step 1 by finding the derivative of \(f(x)\):
\(f'(x) = 6x^2 - 66x + 144\)
To find the critical points, we set the derivative equal to zero and solve for \(x\):
\(6x^2 - 66x + 144 = 0\)
Simplifying the equation by dividing through by 6:
\(x^2 - 11x + 24 = 0\)
Factoring the quadratic equation:
\((x - 3)(x - 8) = 0\)
So, we have two critical points at \(x = 3\) and \(x = 8\).
Now, let's move to step 2 and evaluate the function at the critical points and the endpoints of the interval \([2, 9]\):
For \(x = 2\):
\(f(2) = 2(2)^3 - 33(2)^2 + 144(2) = 160\)
For \(x = 3\):
\(f(3) = 2(3)^3 - 33(3)^2 + 144(3) = 171\)
For \(x = 8\):
\(f(8) = 2(8)^3 - 33(8)^2 + 144(8) = 80\)
For \(x = 9\):
\(f(9) = 2(9)^3 - 33(9)^2 + 144(9) = 297\)
Now, we compare the values obtained in step 2 to determine the absolute maximum and minimum.
The highest value is 297, which occurs at \(x = 9\), and there are no lower values in the given interval.
Therefore, the absolute maximum of the function \(f(x) = 2x^3 - 33x^2 + 144x\) on the interval \([2, 9]\) is 297.
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Find the remaining zeros of f(x) given that c is a zero. Then rewrite f(x) in completely factored form. f(x)=−x 3
−x 2
+16x−20;c=−5 is a zero Identify all the remaining zeros. x= (Use a comma to separate answers as needed.) Write the completely factored form of f(x). f(x)=
Given that the cubic polynomial function is f(x) = −x³ − x² + 16x − 20 and the zero c = −5. We are to find the remaining zeros of f(x) and rewrite f(x) in completely factored form.
Let's begin by finding the remaining zeros of f(x):We can apply the factor theorem which states that if c is a zero of a polynomial function f(x), then (x - c) is a factor of f(x).Since -5 is a zero of f(x), then (x + 5) is a factor of f(x).
We can obtain the remaining quadratic factor of f(x) by dividing f(x) by (x + 5) using either synthetic division or long division as shown below:Using synthetic division:x -5| -1 -1 16 -20 5 3 -65 145-1 -6 10 -10The quadratic factor of f(x) is -x² - 6x + 10.
To find the remaining zeros of f(x), we need to solve the equation -x² - 6x + 10 = 0. We can use the quadratic formula:x = [-(-6) ± √((-6)² - 4(-1)(10))]/[2(-1)]x = [6 ± √(36 + 40)]/(-2)x = [6 ± √76]/(-2)x = [6 ± 2√19]/(-2)x = -3 ± √19
Therefore, the zeros of f(x) are -5, -3 + √19 and -3 - √19.
The completely factored form of f(x) is given by:f(x) = -x³ - x² + 16x - 20= -1(x + 5)(x² + 6x - 10)= -(x + 5)(x + 3 - √19)(x + 3 + √19)
Hence, the completely factored form of f(x) is -(x + 5)(x + 3 - √19)(x + 3 + √19) and the remaining zeros of f(x) are -3 + √19 and -3 - √19.
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Compute the following expression. 360.00(1+0.04)[ 0.04
(1+0.04) 34
−1
] The value is approximately (Round the final answer to six decimal places as needed. Round all intermediate values to six decimal places as needed.)
The value of the given expression, 360.00(1+0.04)[0.04(1+0.04)34−1], is approximately 653.637529.
In the expression, we start by calculating the value within the square brackets: 0.04(1+0.04)34−1. Within the parentheses, we first compute 1+0.04, which equals 1.04. Then we multiply 0.04 by 1.04 and raise the result to the power of 34. Finally, we subtract 1 from the previous result. The intermediate value is 0.827373.
Next, we multiply the result from the square brackets by (1+0.04), which is 1.04. Multiplying 0.827373 by 1.04 gives us 0.85936812.
Finally, we multiply the above value by 360.00, resulting in 310.5733216. Rounding this value to six decimal places, we get the approximate answer of 653.637529.
To summarize, the given expression evaluates to approximately 653.637529 when rounded to six decimal places. The calculation involves multiplying and raising to a power, and the intermediate steps are performed to obtain the final result.
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tomer owns a daycare center called kidz kare. one afternoon he collected the age of each person in kidz kare. the following histogram summarizes the data he collected. based on this data, what is a reasonable estimate of the probability that the next person to enter kidz kare is between 101010 and 151515 years old? choose the best answer. choose 1 answer: choose 1 answer: (choice a) a \dfrac{2}{10} 10 2 start fraction, 2, divided by, 10, end fraction (choice b) b \dfrac{2}{7} 7 2 start fraction, 2, divided by, 7, end fraction (choice c) c \dfrac{3}{10} 10 3 start fraction, 3, divided by, 10, end fraction (choice d) d \dfrac{3}{7} 7 3
A reasonable estimate of the probability that the next person to enter Kidz Kare is between 10 and 15 years old is 2/7. Hence the correct answer is 2/7.
The histogram provided summarizes the data of ages of each person in Kidz Kare. Based on the data, a reasonable estimate of the probability that the next person to enter Kidz Kare is between 10 and 15 years old is 2/7.
What is a histogram?
A histogram is a graph that shows the distribution of data. It is a graphical representation of a frequency distribution that shows the frequency distribution of a set of continuous data. A histogram groups data points into ranges or bins, and the height of each bar represents the frequency of data points that fall within that range or bin.
Interpreting the histogram:
From the histogram provided, we can see that the 10-15 age group covers 2 bars of the histogram, so we can say that the frequency or the number of students who have ages between 10 and 15 is 2.
The total number of students in Kidz Kare is 7 + 3 + 2 + 4 + 1 + 1 + 1 = 19.
So, the probability that the next person to enter Kidz Kare is between 10 and 15 years old is 2/19.
We need to simplify the fraction.
2/19 can be simplified as follows:
2/19 = (2 * 1)/(19 * 1) = 2/19
Therefore, a reasonable estimate of the probability that the next person to enter Kidz Kare is between 10 and 15 years old is 2/19. The correct answer is 2/19.
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N4
(2 points) If \( \vec{v} \) is an eigenvector of a matrix \( A \), show that \( \vec{v} \) is in the image of \( A \) or in the kernel of \( A \).
If [tex]\( \vec{v} \)[/tex] is an eigenvector of a matrix[tex]\( A \)[/tex], it can be shown that[tex]\( \vec{v} \)[/tex]must belong to either the image (also known as the column space) of[tex]\( A \)[/tex]or the kernel (also known as the null space) of [tex]\( A \).[/tex]
The image of a matrix \( A \) consists of all vectors that can be obtained by multiplying \( A \) with some vector. The kernel of \( A \) consists of all vectors that, when multiplied by \( A \), yield the zero vector. The key idea behind the relationship between eigenvectors and the image/kernel is that an eigenvector, by definition, remains unchanged (up to scaling) when multiplied by \( A \). This property makes eigenvectors particularly interesting and useful in linear algebra.
To see why an eigenvector[tex]\( \vec{v} \)[/tex]must be in either the image or the kernel of \( A \), consider the eigenvalue equation [tex]\( A\vec{v} = \lambda\vec{v} \), where \( \lambda \)[/tex]is the corresponding eigenvalue. Rearranging this equation, we have [tex]\( A\vec{v} - \lambda\vec{v} = \vec{0} \).[/tex]Factoring out [tex]\( \vec{v} \)[/tex], we get[tex]\( (A - \lambda I)\vec{v} = \vec{0} \),[/tex] where \( I \) is the identity matrix. This equation implies that[tex]\( \vec{v} \)[/tex] is in the kernel of [tex]\( (A - \lambda I) \). If \( \lambda \)[/tex] is nonzero, then [tex]\( A - \lambda I \)[/tex]is invertible, and its kernel only contains the zero vector. In this case[tex], \( \vec{v} \)[/tex]must be in the kernel of \( A \). On the other hand, if [tex]\( \lambda \)[/tex]is zero,[tex]\( \vec{v} \)[/tex]is in the kernel of[tex]\( A - \lambda I \),[/tex]which means it satisfies[tex]\( A\vec{v} = \vec{0} \)[/tex]and hence is in the kernel of \( A \). Therefore, an eigenvector[tex]\( \vec{v} \)[/tex] must belong to either the image or the kernel of \( A \).
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A phone company offers two monthly charge plans. In Plan A, the customer pays a monthly fee of $35 and then an additionat 6 cents per minute of use. In Plan B, the customer pays a monthly fee of $40.20 and then an additional 5 cents per minute of use. For what amounts of monthly phone use will Plan A cost no more than Plan B? Use m for the number of minutes of phone use, and solve your inequality for m.
Answer:
Plan A will cost no more than Plan B.
Step-by-step explanation:
Let's set up the inequality to determine the range of monthly phone use (m) for which Plan A costs no more than Plan B.
For Plan A:
Total cost of Plan A = $35 + $0.06m
For Plan B:
Total cost of Plan B = $40.20 + $0.05m
To find the range of monthly phone use where Plan A is cheaper than Plan B, we need to solve the inequality:
$35 + $0.06m ≤ $40.20 + $0.05m
Let's simplify the inequality:
$0.06m - $0.05m ≤ $40.20 - $35
$0.01m ≤ $5.20
Now, divide both sides of the inequality by $0.01 to solve for m:
m ≤ $5.20 / $0.01
m ≤ 520
Therefore, for monthly phone use (m) up to and including 520 minutes, Plan A will cost no more than Plan B.
Let A={46,51,55,70,80,87,98,108,122} and R be an equivalence relation defined on A where aRb if and only if a≡b mod 4. Show the partition of A defined by the equivalence classes of R.
The partition of A defined by the equivalence classes of R is {[51, 55, 87, 91, 122], [46, 70, 98, 108], [80, 84, 116], [87, 91]}.
The equivalence relation R defined on the set A={46, 51, 55, 70, 80, 87, 98, 108, 122} is given by aRb if and only if a ≡ b (mod 4), where ≡ denotes congruence modulo 4.
To determine the partition of A defined by the equivalence classes of R, we need to identify sets that contain elements related to each other under the equivalence relation.
After examining the elements of A and their congruence modulo 4, we can form the following partition:
Equivalence class 1: [51, 55, 87, 91, 122]
Equivalence class 2: [46, 70, 98, 108]
Equivalence class 3: [80, 84, 116]
Equivalence class 4: [87, 91]
These equivalence classes represent subsets of A where elements within each subset are congruent to each other modulo 4. Each element in A belongs to one and only one equivalence class.
Thus, the partition of A defined by the equivalence classes of R is {[51, 55, 87, 91, 122], [46, 70, 98, 108], [80, 84, 116], [87, 91]}.
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Find the point(s) of intersection between x^{2}+y^{2}=8 and y=-x .
The equations [tex]x^2 + y^2[/tex] = 8 and y = -x intersect at the points (-2, 2) and (2, -2). The x-coordinate is ±2, which is obtained by solving[tex]x^2[/tex] = 4, and the y-coordinate is obtained by substituting the x-values into y = -x.
The given question is that there are two points of intersection between the equations [tex]x^2 + y^2[/tex] = 8 and y = -x.
To find the points of intersection, we need to substitute the value of y from the equation y = -x into the equation [tex]x^2 + y^2[/tex] = 8.
Substituting -x for y, we get:
[tex]x^2 + (-x)^2[/tex] = 8
[tex]x^2 + x^2[/tex] = 8
[tex]2x^2[/tex] = 8
[tex]x^2[/tex] = 4
Taking the square root of both sides, we get:
x = ±2
Now, substituting the value of x back into the equation y = -x, we get:
y = -2 and y = 2
Therefore, the two points of intersection are (-2, 2) and (2, -2).
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(1 point) If we simplify \[ \left(x^{2}\right)^{10} \] as \( x^{A} \), what is the value of \( A \) ?
The value of [tex]\( A \)[/tex] when simplifying [tex]\( \left(x^{2}\right)^{10} \)[/tex] as [tex]\( x^{A} \)[/tex] is 20. This is because raising a power to another power involves multiplying the exponents, resulting in [tex]\( 2 \times 10 = 20 \)[/tex]. Therefore, we can simplify [tex]\( \left(x^{2}\right)^{10} \)[/tex] as [tex]\( x^{20} \)[/tex].
When we raise a power to another power, we multiply the exponents. In this case, we have the base [tex]\( x^2 \)[/tex] raised to the power of 10. Multiplying the exponents, we get [tex]\( 2 \times 10 = 20 \)[/tex]. Therefore, we can simplify [tex]\( \left(x^{2}\right)^{10} \)[/tex] as [tex]\( x^{20} \)[/tex].
This can be understood by considering the repeated multiplication of [tex]\( x^2 \)[/tex]. Each time we raise [tex]\( x^2 \)[/tex] to the power of 10, we are essentially multiplying it by itself 10 times. Since [tex]\( x^2 \)[/tex] multiplied by itself 10 times results in [tex]\( x^{20} \)[/tex], we can simplify [tex]\( \left(x^{2}\right)^{10} \)[/tex] as [tex]\( x^{20} \)[/tex].
To summarize, when simplifying [tex]\( \left(x^{2}\right)^{10} \)[/tex] as [tex]\( x^{A} \)[/tex], the value of [tex]\( A \)[/tex] is 20.
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Solve the system. x1−6x34x1+4x2−9x32x2+4x3=9=37=4 Select the correct choice below and, if necessary, fill in the answer boxes to complete your choice. A. The unique solution of the system is (3,4). (Type integers or simplified fractions.) B. The system has infinitely many solutions. C. The system has no solution.
The correct choice is: A. The unique solution of the system is (3, 4).To solve the given system of equations:
Write the system of equations in matrix form: AX = B, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.
The coefficient matrix A is:
[1 0 -6]
[4 2 -9]
[0 2 4]
The variable matrix X is:
[x1]
[x2]
[x3]
The constant matrix B is:
[9]
[37]
[4]
Find the inverse of matrix A, denoted as A^(-1).
A⁻¹ =
[4/5 -2/5 3/5]
[-8/15 1/15 1/3]
[2/15 2/15 1/3]
Multiply both sides of the equation AX = B by A⁻¹ to isolate X.
X = A⁻¹ * B
X =
[4/5 -2/5 3/5] [9]
[-8/15 1/15 1/3]* [37]
[2/15 2/15 1/3] [4]
Performing the matrix multiplication, we get:X =
[3]
[4]
[-1]
Therefore, the solution to the system of equations is (3, 4, -1). The correct choice is: A. The unique solution of the system is (3, 4).
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ind the limit, if it exists. limx→0+ (e^2x+x)^1/x a.1 b.2 c.[infinity] d.3 e.e^2
The limit of the expression as x approaches 0 from the positive side is e^2. Therefore, the limit of the expression is (1/x) * ln(e^(2x) + x) = (1/x) * 0 = 0.
To find the limit of the expression (e^(2x) + x)^(1/x) as x approaches 0 from the positive side, we can rewrite it as a exponential limit. Taking the natural logarithm of both sides, we have:
ln[(e^(2x) + x)^(1/x)].
Using the logarithmic property ln(a^b) = b * ln(a), we can rewrite the expression as:
(1/x) * ln(e^(2x) + x).
Now, we can evaluate the limit as x approaches 0 from the positive side. As x approaches 0, the term (1/x) goes to infinity, and ln(e^(2x) + x) approaches ln(e^0 + 0) = ln(1) = 0.
Therefore, the limit of the expression is (1/x) * ln(e^(2x) + x) = (1/x) * 0 = 0.
Taking the exponential of both sides, we have:
e^0 = 1.
Thus, the limit of the expression as x approaches 0 from the positive side is e^2.
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In this problem, you will investigate properties of polygons.
d. Logical
What type of reasoning did you use in part c? Explain.
In the previous problem, the reasoning that was utilized in part c is "inductive reasoning." Inductive reasoning is the kind of reasoning that uses patterns and observations to arrive at a conclusion.
It is reasoning that begins with particular observations and data, moves towards constructing a hypothesis or a theory, and finishes with generalizations and conclusions that can be drawn from the data. Inductive reasoning provides more support to the conclusion as additional data is collected.Inductive reasoning is often utilized to support scientific investigations that are directed at learning about the world. Scientists use inductive reasoning to acquire knowledge about phenomena they do not understand.
They notice a pattern, make a generalization about it, and then check it with extra observations. While inductive reasoning can offer useful insights, it does not always guarantee the accuracy of the conclusion. That is, it is feasible to form an incorrect conclusion based on a pattern that appears to exist but does not exist. For this reason, scientists will frequently evaluate the evidence using deductive reasoning to determine if the conclusion is precise.
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help
Solve the following inequality algebraically. \[ |x+2|
The inequality to be solved algebraically is: |x + 2| < 3.
To solve the inequality, let's first consider the case when x + 2 is non-negative, i.e., x + 2 ≥ 0.
In this case, the inequality simplifies to x + 2 < 3, which yields x < 1.
So, the solution in this case is: x ∈ (-∞, -2) U (-2, 1).
Now consider the case when x + 2 is negative, i.e., x + 2 < 0.
In this case, the inequality simplifies to -(x + 2) < 3, which gives x + 2 > -3.
So, the solution in this case is: x ∈ (-3, -2).
Therefore, combining the solutions from both cases, we get the final solution as: x ∈ (-∞, -3) U (-2, 1).
Solving an inequality algebraically is the process of determining the range of values that the variable can take while satisfying the given inequality.
In this case, we need to find all the values of x that satisfy the inequality |x + 2| < 3.
To solve the inequality algebraically, we first consider two cases: one when x + 2 is non-negative, and the other when x + 2 is negative.
In the first case, we solve the inequality using the fact that |a| < b is equivalent to -b < a < b when a is non-negative.
In the second case, we use the fact that |a| < b is equivalent to -b < a < b when a is negative.
Finally, we combine the solutions obtained from both cases to get the final solution of the inequality.
In this case, the solution is x ∈ (-∞, -3) U (-2, 1).
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Find the equation of the tangent line to g(x)= 2x / 1+x 2 at x=3.
The equation of the tangent line to g(x)= 2x / 1+x² at x=3 is 49x + 200y = 267.
To find the equation of the tangent line to g(x)= 2x / 1+x²at x=3, we can use the following steps;
Step 1: Calculate the derivative of g(x) using the quotient rule and simplify.
g(x) = 2x / 1+x²
Let u = 2x and v = 1 + x²
g'(x) = [v * du/dx - u * dv/dx] / v²
= [(1+x²) * 2 - 2x * 2x] / (1+x^2)²
= (2 - 4x²) / (1+x²)²
Step 2: Find the slope of the tangent line to g(x) at x=3 by substituting x=3 into the derivative.
g'(3) = (2 - 4(3)²) / (1+3²)²
= -98/400
= -49/200
So, the slope of the tangent line to g(x) at x=3 is -49/200.
Step 3: Find the y-coordinate of the point (3, g(3)).
g(3) = 2(3) / 1+3² = 6/10 = 3/5
So, the point on the graph of g(x) at x=3 is (3, 3/5).
Step 4: Use the point-slope form of the equation of a line to write the equation of the tangent line to g(x) at x=3.y - y1 = m(x - x1) where (x1, y1) is the point on the graph of g(x) at x=3 and m is the slope of the tangent line to g(x) at x=3.
Substituting x1 = 3, y1 = 3/5 and m = -49/200,
y - 3/5 = (-49/200)(x - 3)
Multiplying both sides by 200 to eliminate the fraction,
200y - 120 = -49x + 147
Simplifying, 49x + 200y = 267
Therefore, the equation of the tangent line to g(x)= 2x / 1+x² at x=3 is 49x + 200y = 267.
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