The arbitrary constant is eliminated when we take the derivative of the equation [tex]y = Ax^5 + Bx^3[/tex], resulting in [tex]dy/dx = 5Ax^4 + 3Bx^2.[/tex]
To eliminate the arbitrary constant from the equation [tex]y = Ax^5 + Bx^3[/tex], we can take the derivative of both sides with respect to x.
[tex]d/dx (y) = d/dx (Ax^5 + Bx^3)\\dy/dx = 5Ax^4 + 3Bx^2[/tex]
Now, we have the derivative of y with respect to x. The arbitrary constant is eliminated in this process.
To know more about arbitrary constant,
https://brainly.com/question/33151029
#SPJ11
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))).
Learn more about asymptotic relationship here:
brainly.com/question/32388884
#SPJ11
Given are three simple linear equations in the format of y=mx+b. Equation 1: y=25,105+0.69x Equation 2:y=7,378+1.41x Equation 3:y=12.509+0.92x Instructions 1. Plot and label all equations 1. 2 and 3 on the same graph paper. 2. The graph must show how these equations intersect with each other if they do. Label each equation (8 pts.). 3. Compute each Interception point (coordinate). On the graph label each interception point with its coordinate (8 pts.) 4. Upload your graph in a pdf format (zero point for uploading a non-pdf file) by clicking in the text box below and selecting the paper dip symbol.
According to given information, the graph plotting and uploading steps are given below.
Given linear equations are: y = 25,105 + 0.69xy = 7,378 + 1.41xy = 12.509 + 0.92x
To plot and label the given linear equations, follow these steps:
Draw a graph on a graph paper with x and y-axis.
Draw the line for each linear equation by identifying two points on the line and connecting them using a straight line. To find two points on the line, substitute any value of x and solve for y using the given equation. This will give you one point on the line.
Now, substitute a different value of x and solve for y.
This will give you another point on the line.
Label each line with the equation it represents.
Find the point of intersection of each pair of lines by solving the system of equations formed by those two lines. You can do this by substituting one equation into the other to find the value of x.
Then, substitute this value of x back into either equation to find the value of y. This will give you the point of intersection of those two lines.
Label each point of intersection with its coordinates.
Once you have drawn all three lines and identified their points of intersection, your graph is complete.
Finally, upload your graph in pdf format.
To know more about coordinates, visit:
https://brainly.com/question/32836021
#SPJ11
Let X 1
,…,X n
∼Beta(θ,2). Show that T=∏ i=1
n
X i
is a sufficient statistic for θ. Note: You may simplify the pdf before you proceed f(x∣θ)= Γ(θ)Γ(2)
Γ(θ+2)
x θ−1
(1−x) 2−1
To show that the product statistic T = ∏ᵢ₌₁ⁿ Xᵢ is sufficient for θ, we need to demonstrate that the conditional distribution of the sample given T does not depend on θ.
Given that X₁, X₂, ..., Xₙ are i.i.d. random variables with a Beta distribution Beta(θ, 2), we can express the joint probability density function (pdf) of the sample as:
f(x₁, x₂, ..., xₙ | θ) = ∏ᵢ₌₁ⁿ f(xᵢ | θ)
= ∏ᵢ₌₁ⁿ [Γ(θ)Γ(2) / Γ(θ + 2)] * xᵢ^(θ - 1) * (1 - xᵢ)^(2 - 1)
= [Γ(θ)Γ(2) / Γ(θ + 2)]ⁿ * ∏ᵢ₌₁ⁿ xᵢ^(θ - 1) * (1 - xᵢ)
To proceed, let's rewrite the joint pdf in terms of the product statistic T:
f(x₁, x₂, ..., xₙ | θ) = [Γ(θ)Γ(2) / Γ(θ + 2)]ⁿ * T^(θ - 1) * (1 - T)^(2n - θ)
Now, let's factorize the joint pdf into two parts, one depending on the data and the other on the parameter:
f(x₁, x₂, ..., xₙ | θ) = g(T, θ) * h(x₁, x₂, ..., xₙ)
where g(T, θ) = [Γ(θ)Γ(2) / Γ(θ + 2)]ⁿ * T^(θ - 1) * (1 - T)^(2n - θ) and h(x₁, x₂, ..., xₙ) = 1.
The factorization shows that the joint pdf can be separated into a function of T, which depends on the parameter θ, and a function of the data x₁, x₂, ..., xₙ. Since the factorization does not depend on the specific values of x₁, x₂, ..., xₙ, we can conclude that the product statistic T = ∏ᵢ₌₁ⁿ Xᵢ is a sufficient statistic for θ.
To know more about Beta distribution, visit:
https://brainly.com/question/32657045
#SPJ11
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.
Learn more on ways to select on https://brainly.com/question/13773260
#SPJ4
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.
To know more about equation,
https://brainly.com/question/32749704
#SPJ11
The simplest measure of dispersion in a data set is the: A. Range B. Standard deviation C. Variance D. Inter quartile range
The simplest measure of dispersion in a data set is the range. This is option A.The answer is the range. A range can be defined as the difference between the largest and smallest observations in a data set, making it the simplest measure of dispersion in a data set.
The range can be calculated as: Range = Maximum observation - Minimum observation.
Range: the range is the simplest measure of dispersion that is the difference between the largest and the smallest observation in a data set. To determine the range, subtract the minimum value from the maximum value. Standard deviation: the standard deviation is the most commonly used measure of dispersion because it considers each observation and is influenced by the entire data set.
Variance: the variance is similar to the standard deviation but more complicated. It gives a weight to the difference between each value and the mean.
Interquartile range: The difference between the third and the first quartile values of a data set is known as the interquartile range. It's a measure of the spread of the middle half of the data. The interquartile range is less vulnerable to outliers than the range. However, the simplest measure of dispersion in a data set is the range, which is the difference between the largest and smallest observations in a data set.
The simplest measure of dispersion is the range. The range is calculated by subtracting the minimum value from the maximum value. The range is useful for determining the distance between the two extreme values of a data set.
To know more about Standard deviation visit:
brainly.com/question/13498201
#SPJ11
For each system, determine whether it has a unique solution (in this case, find the solution), infinitely many solutions, or no solutions. a.
-2x+2y= 10
-4x+4y = 20
A. Infinitely many solutions
B. Unique solution: -0, -0
C. No solutions
D. Unique solution: z=-5, y=0
E. Unique solution: -
F. None of the above
The solution to the system of equations is \boxed{\textbf{(D) } \text{Unique solution: }x=-5, y=0}.
Let us solve the following system of equations: \begin{aligned}-2x+2y &= 10\\-4x+4y &= 20\end{aligned}$$
We can simplify the second equation by dividing both sides by 4.
This will give us the same equation as the first. \begin{aligned}-2x+2y &= 10\\-x+y &= 5\end{aligned}
This system of equations can be solved by adding the equations together.
-2x + 2y + (-x + y) = 10 + 5-3x + 3y = 15 -3(x - y) = 15 x - y = -5
Therefore, the system of equations has a unique solution. The solution is \begin{aligned}x - y &= -5\\x &= -5 + y\end{aligned}
Therefore, we can use either equation in the original system of equations to solve for y-2x+2y= 10-2(-5 + y) + 2y = 10, 10 - 2y + 2y = 10, 0 = 0
Since 0 = 0, the value of y does not matter. We can choose any value for y and solve for x. For example, if we let y = 0, then x - y = -5x - 0 = -5 x = -5
Therefore, the solution to the system of equations is \boxed{\textbf{(D) } \text{Unique solution: }x=-5, y=0}.
To know more about system of equations visit:
brainly.com/question/33289088
#SPJ11
Assume that f is a one-to-one function. If f(4)=−7, find f−1(−7)
Given that f is a one-to-one function and f(4) = -7. We need to find f⁻¹(-7). The definition of one-to-one function f is a one-to-one function, it means that each input has a unique output. In other words, there is a one-to-one correspondence between the domain and range of the function. It also means that for each output of the function, there is one and only one input. Let us denote f⁻¹ as the inverse of f and x as f⁻¹(y). Now we can represent the given function as: f(x) = -7Let y = f(x) and x = f⁻¹(y) Now substituting f⁻¹(y) in place of x, we get: f(f⁻¹(y)) = -7Since f(f⁻¹(y)) = y We get: y = -7Therefore, f⁻¹(-7) = 4 Hence, f⁻¹(-7) = 4.
To learn more about one-to-one function:https://brainly.com/question/28911089
#SPJ11
$4.845 is invested, part at 13% and the rest at 7%. If the interest eamed from the amount invested at 13 of eiceeds the interest earned from the amount invested at 7% by $188.65, how much is irvested at each rate? (Round to two decimal places if necessary)
$2634 is invested at 13% interest rate and $2211 ($4845-$2634) is invested at 7% interest rate. Amount invested at 13% = $2634Amount invested at 7% = $2211
Let's start the solution of the given problem below; Let X be the amount invested at 13% interest rate and the remaining amount, which is invested at 7% interest rate. Then, Interest earned on the amount invested at 13% interest rate will be 0.13X.Interest earned on the amount invested at 7% interest rate will be 0.07(4845 - X) = 338.15 - 0.07X.
The interest earned from the amount invested at 13% exceeds the interest earned from the amount invested at 7% by $188.65, this can be written in an equation as;0.13X - (338.15 - 0.07X) = 188.65 0.13X - 338.15 + 0.07X = 188.65 0.20X = 526.80 X = 2634. Thus, $2634 is invested at 13% interest rate and $2211 ($4845-$2634) is invested at 7% interest rate. Answer: Amount invested at 13% = $2634Amount invested at 7% = $2211.
Let's learn more about interest:
https://brainly.com/question/25720319
#SPJ11
Use the defining formula, the computation formula, or a calculator to compute s. (Round your answer to four decimal places.) s= (b) Multiply each data value by 3 to obtain the new data set 33,45,51,33,24. Compute s. (Round your answer to four decimal places.) s= (c) Compare the results of parts (a) and (b). In general, how does the standard deviation change if each data value is multiplied by a constant c? Multiplying each data value by the same constant c results in the standard deviation being ∣c∣ times smaller. Multiplying each data value by the same constant c results in the standard deviation being ∣c∣ times as ∣arge. Multiplying each data value by the same constant c results in the standard deviation remaining the same. Multiplying each data value by the same constant c results in the standard deviation increasing by c units. (d) You recorded the weekly distances you bicycled in miles and computed the standard deviation to be s=4 miles. Your friend wants to know the standard deviation in kilometers. Do you need to redo all the calculations? Yes No Given 1 mile ≈1.6 kilometers, what is the standard deviation in kilometers? (Enter your answer to two decimal places.) s= km In this problem, we explore the effect on the standard deviation of multiplying each data value in a data set by the same constant. Consider the data set 11,15,17,11,8. (a) Use the defining formula, the computation formula, or a calculator to compute s. (Round your answer to four decimal places.) s= (b) Multiply each data value by 3 to obtain the new data set 33,45,51,33,24. Compute s. (Round your answer to four decimal places.) s= (c) Compare the results of parts (a) and (b). In general, how does the standard deviation change if each data value is multiplied by a constant c? Multiplying each data value by the same constant c results in the standard deviation being ∣c∣ times smaller. Multiplying each data value by the same constant c results in the standard deviation being ∣c∣ times as large. Multiplying each data value by the same constant c results in the standard deviation remaining the same. Multiplying each data value by the same constant c results in the standard deviation increasing by c units. (d) You recorded the weekly distances you bicycled in miles and computed the standard deviation to be s=4 miles. Your friend wants to know the standard deviation in kilometers. Do you need to redo all the calculations? Yes No Given 1 mile ≈1.6 kilometers, what is the standard deviation in kilometers? (Enter your answer to two decimal places.) s= km
a) To compute s for the given data set, we use the formula, where μ is the mean and N is the total number of data points.
b) If we multiply each data value by 3, the new data set will be as follows:33, 45, 51, 33, 24
The formula to compute s for this data set is similar to the one used in part a. We have
c) We can observe that the standard deviation changes if each data value is multiplied by a constant c.
If we multiply each data value by the same constant c, the standard deviation is |c| times larger.
For example, if we multiply each data value by 3, the standard deviation becomes 3 times larger than the original standard deviation.
Know more about data set here:
https://brainly.com/question/30154121
#SPJ11
At least one of the answers above is NOT correct. (1 point ) Find the quotient and remainder using synthetic division for (x^(3)-12x^(2)+34x-12)/(x-4) The quotient is The remainder is Note: You can ea
Therefore, the quotient is [tex]x^2 + 4x + 66[/tex], and the remainder is 252.
To find the quotient and remainder using synthetic division for the polynomial division of [tex](x^3 - 12x^2 + 34x - 12)[/tex] by (x - 4), we follow these steps:
Set up the synthetic division table, representing the divisor (x - 4) and the coefficients of the dividend [tex](x^3 - 12x^2 + 34x - 12)[/tex]:
Bring down the first coefficient of the dividend (1) into the leftmost slot of the synthetic division table:
Multiply the divisor (4) by the value in the result row (1), and write the product (4) below the second coefficient of the dividend (-12). Add the two numbers (-12 + 4 = -8) and write the sum in the second slot of the result row:
Repeat the process, multiplying the divisor (4) by the new value in the result row (-8), and write the product (32) below the third coefficient of the dividend (34). Add the two numbers (34 + 32 = 66) and write the sum in the third slot of the result row:
Multiply the divisor (4) by the new value in the result row (66), and write the product (264) below the fourth coefficient of the dividend (-12). Add the two numbers (-12 + 264 = 252) and write the sum in the fourth slot of the result row:
The numbers in the result row, from left to right, represent the coefficients of the quotient. In this case, the quotient is: [tex]x^2 + 4x + 66.[/tex]
The number in the bottom right corner of the synthetic division table represents the remainder. In this case, the remainder is 252.
To know more about quotient,
https://brainly.com/question/29248338
#SPJ11
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]
Learn more about triple integral at:
https://brainly.com/question/30404807
#SPJ4
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]
The reduced row-echelon fo of the augmented matrix for a system of linear equations with variables x1,…,x5 is given below. Deteine the solutions for the sys and enter them below. ⎣⎡1000100015−52−3−125−5−5⎦⎤ If the system has infinitely many solutions, select "The system has at least one solution". Your answer may use expressions involving the parameters r, s, and f. The system has no solutions
The given matrix represents the augmented matrix of a system of linear equations. To determine the solutions of the system, we need to analyze the row-echelon form. The given matrix is: ⎣⎡1000100015−52−3−125−5−5⎦⎤We can now convert this matrix to row-echelon form, then reduced row-echelon form to get the solutions of the system. To convert to row-echelon form, we can use Gaussian elimination and get the following matrix. ⎣⎡1000100010−52−3−12000⎦⎤We can then convert this matrix to reduced row-echelon form to get the solutions. ⎣⎡1000100010−520−130000⎦⎤The last non-zero row corresponds to the equation 0=1, which is impossible and therefore the system has no solutions. Therefore, the correct option is "The system has no solutions".
To learn more about echelon form:https://brainly.com/question/28968080
#SPJ11
PLS ANSWER 100 POINTS
A coffee mixture has beans that sell for $0.52 a pound and beans that sell for $0.28. If 130 pounds of beans create a mixture worth $0.64 a pound, how much of each bean is used? Model the scenario then solve it. Then, in two or more sentences explain whether your solution is or is not reasonable.
Answer:
To solve this problem, we can use a system of two equations with two unknowns. Let x be the number of pounds of beans that sell for $0.52 per pound, and let y be the number of pounds of beans that sell for $0.28 per pound. We can write:
x + y = 130 (the total weight of beans is 130 pounds)
0.52x + 0.28y = 0.64(130) (the value of the mixture is $0.64 per pound)
Solving this system of equations, we get x = 50 and y = 80, which means that 50 pounds of $0.52-per-pound beans and 80 pounds of $0.28-per-pound beans are used in the mixture.
This solution is reasonable because it satisfies both equations and makes sense in the context of the problem. The sum of the weights of the two types of beans is 130 pounds, which is the total weight of the mixture, and the value of the mixture is $0.64 per pound, which is the desired value. The amount of the cheaper beans is higher than the amount of the more expensive beans, which is also reasonable since the cheaper beans contribute more to the total weight of the mixture.
i) Are the following equalities generally valid? A ∪ (B \ C) = (A ∪ B) \ (A ∪ C)
A ∩ (B \ C) = (A ∩ B) \ (A ∩ C)
Give a counterexample or prove the argument
ii) Give an example of a set A containing at least one element that fulfills the condition
if x ∈ A so {x} ∈ A
1. The equalities are not generally valid.
2. 0 is an element of A, and {0} is also an element of A since it is a singleton set containing 0.
i) The equalities A ∪ (B \ C) = (A ∪ B) \ (A ∪ C) and A ∩ (B \ C) = (A ∩ B) \ (A ∩ C) are not generally valid.
Counterexample for A ∪ (B \ C) = (A ∪ B) \ (A ∪ C):
Let A = {1, 2}, B = {2, 3}, and C = {1, 3}.
A ∪ (B \ C) = {1, 2} ∪ {2} = {1, 2}
(A ∪ B) \ (A ∪ C) = ({1, 2} ∪ {2, 3}) \ ({1, 2} ∪ {1, 3}) = {1, 2, 3} \ {1, 2} = {3}
Since {1, 2} is not equal to {3}, the equality A ∪ (B \ C) = (A ∪ B) \ (A ∪ C) does not hold in this case.
Counterexample for A ∩ (B \ C) = (A ∩ B) \ (A ∩ C):
Let A = {1, 2}, B = {2, 3}, and C = {1, 3}.
A ∩ (B \ C) = {1, 2} ∩ {2} = {2}
(A ∩ B) \ (A ∩ C) = ({1, 2} ∩ {2, 3}) \ ({1, 2} ∩ {1, 3}) = {2} \ {1, 2} = {}
Since {2} is not equal to {}, the equality A ∩ (B \ C) = (A ∩ B) \ (A ∩ C) does not hold in this case.
Therefore, the equalities are not generally valid.
ii) An example of a set A containing at least one element that fulfills the condition if x ∈ A, then {x} ∈ A is:
A = {0, {0}}
In this case, 0 is an element of A, and {0} is also an element of A since it is a singleton set containing 0.
Learn more about Elements from
https://brainly.com/question/25916838
#SPJ11
Let f(x)=3x+5
Find f'(x)
a)none of these
b) f'(x) = 5
c) f'(x)=3
d) f'(x) = x
Answer:
f(x) = 3x + 5, so f'(x) = 3.
The correct answer is c.
an automobile manufacturer buys computer chips from a supplier. the supplier sends a shipment containing 5% defective chips. each chip chosen from this shipment has a probability of 0.05% of being defective, and each automobile uses 12 chips selected independently. what is the probability that all 12 chips in a car will work properly
The probability that all 12 chips in a car will work properly is approximately 0.9888, or 98.88%.
To determine the probability that all 12 chips in a car will work properly, we need to calculate the probability of selecting a non-defective chip and then raise it to the power of 12.
we are given that each chip has a 0.05% probability of being defective, the probability of selecting a non-defective chip is 1 - 0.05% = 99.95%.
To determine the probability that all 12 chips in a car will work properly, we raise this probability to the power of 12:
P(all 12 chips work properly) = [tex](99.95)^{12}[/tex]
P(all 12 chips work properly) = [tex](0.9995)^{12}[/tex] ≈ 0.9888
Therefore, the probability that all 12 chips in a car will work properly is approximately 0.9888, or 98.88%.
This means that there is a 98.88% chance that none of the 12 chips in a car will be defective.
Learn more about probability Visit :
brainly.com/question/13604758
#SPJ4
The probability that someone is wearing sunglasses and a hat is 0.25 The probability that someone is wearing a hat is 0.4 The probability that someone is wearing sunglasses is 0.5 Using the probability multiplication rule, find the probability that someone is wearing a hat given that they are wearin
To find the probability that someone is wearing a hat given that they are wearing sunglasses, we can use the probability multiplication rule, also known as Bayes' theorem.
Let's denote:
A = event of wearing a hat
B = event of wearing sunglasses
According to the given information:
P(A and B) = 0.25 (the probability that someone is wearing both sunglasses and a hat)
P(A) = 0.4 (the probability that someone is wearing a hat)
P(B) = 0.5 (the probability that someone is wearing sunglasses)
Using Bayes' theorem, the formula is:
P(A|B) = P(A and B) / P(B)
Substituting the given probabilities:
P(A|B) = 0.25 / 0.5
P(A|B) = 0.5
Therefore, the probability that someone is wearing a hat given that they are wearing sunglasses is 0.5, or 50%.
To learn more about Bayes' theorem:https://brainly.com/question/14989160
#SPJ11
Create a scatterplot for the data in the Weight and the City MPG columns. Paste it here. a) Using Stat Disk, calculate the linear correlation between the data in the Weight and City MPG columns. Paste your results in your Word document. b) Explain the mathematical relationship between Weight and City MPG based on the linear correlation coefficient. Be certain to include comments about the magnitude and the direction of the correlation. c) Compare and contrast the correlations for weight and braking distance with that of weight and city MPG. How are they similar and how are they different?
The scatterplot for the data in the Weight and the City MPG columns is: The calculation of linear correlation between the data in the Weight and City MPG columns with Stat Disk is shown below;Linear Correlation Coefficient = -0.812
The mathematical relationship between Weight and City MPG is that there is a strong negative correlation between the two variables. When the weight increases, the City MPG decreases, and vice versa. The correlation coefficient is -0.812, which indicates a strong correlation, and the negative sign represents the inverse relationship. If the weight of a car increases, its fuel efficiency will decrease, and vice versa. The magnitude of correlation is moderate to high. The higher the magnitude, the stronger the correlation between the two variables. The direction of the correlation is negative, which implies that the variables move in the opposite direction. When one variable decreases, the other increases, and vice versa. The correlation between weight and braking distance is positive, and the correlation between weight and City MPG is negative. The positive correlation between weight and braking distance indicates that as the weight of a car increases, the braking distance also increases. There is a negative correlation between weight and City MPG, which means that the fuel efficiency decreases as the weight of a car increases. As one variable increases, the other decreases in weight and City MPG, while the opposite is true for weight and braking distance.
In conclusion, we can infer that there is a strong negative correlation between weight and City MPG. The higher the weight of a car, the lower its fuel efficiency, and vice versa. There is a moderate to high magnitude of correlation and an inverse relationship between the two variables. The comparison of weight and braking distance with that of weight and City MPG revealed that there are differences in their correlation coefficients and directions.
To learn more about Linear Correlation Coefficient visit:
brainly.com/question/24881420
#SPJ11
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.
Learn more about current
brainly.com/question/31315986
#SPJ11
Shota built a time travel machine, but he can't control the duration of his trip. Each time he uses a machine he has a 0.8 probability of staying in the alternative time for more than an hour. During the first year of testing, Shots uses his machine 20 times. Assuming that each trip is equally likely to last for more than an hour, what is the probability that at least one trip will last less than an hour? Round your answer to the nearest hundredth. P(at least one < 1 hour) =
The probability that at least one trip will last less than an hour is approximately 0.99. when rounded to the nearest hundredth.
Given,
Each trip has a probability of lasting more than an hour = 0.8
The probability of any individual trip lasting less than an hour is
1 - 0.8 = 0.2.
Since each trip is assumed to be independent and equally likely, the probability of all 20 trips lasting more than an hour is
[tex](0.8)^{20}[/tex]= 0.011529215.
Therefore, the probability of at least one trip lasting less than an hour
1- 0.011529215 = 0.988470785.
Rounded to the nearest hundredth, the probability is approximately 0.99.
Learn more about probability here:
https://brainly.com/question/30140200
#SPJ4
find the standard for, of equation of am ellipse with center at the orgim major axis on the y axix a=10and b=7
The standard equation of an ellipse with center at the origin, major axis on the y-axis, and a = 10 and b = 7 is
x^2/49 + y^2/100 = 1
The standard form of the equation of an ellipse with center at the origin is
x^2/a^2 + y^2/b^2 = 1.
Since the major axis is on the y-axis, the larger value, which is 10, is assigned to b and the smaller value, which is 7, is assigned to a.
Thus, the equation is:
x^2/7^2 + y^2/10^2 = 1
Multiplying both sides by 7^2 x 10^2, we obtain:
100x^2 + 49y^2 = 4900
Dividing both sides by 4900, we get:
x^2/49 + y^2/100 = 1
Therefore, the standard form of the equation of the given ellipse is x^2/49 + y^2/100 = 1.
To know more about ellipse visit:
https://brainly.com/question/20393030
#SPJ11
pls
ans 3
Eliminate the arbitrary constant C. y=x^{2}+C e^{-x} \[ y^{\prime}-y=2 x-x^{2} \] \[ y^{\prime}+x y=x^{3}+2 x \] \[ x y^{\prime}+y=3 x^{2} \] \[ y^{\prime}+y=x^{2}+2 x \]
What is the best descr
The particular solution to the differential equation with the initial condition y(0) = 1 is:
(1/2)x^2 + ln|y| = 0
ln|y| = -(1/2)x^2
|y| = e^(-(1/2)x^2)
y = ±e^(-(1/2)x^2)
The differential equation given is:
y = x^2 + Ce^(-x) ...(1)
We need to eliminate the arbitrary constant C from equation (1) and obtain a particular solution.
To do this, we differentiate both sides of equation (1) with respect to x:
dy/dx = 2x - Ce^(-x) ...(2)
Substituting equation (1) into the given differential equations, we get:
y' - y = 2x - x^2
Substituting y = x^2 + Ce^(-x), and y' = 2x - Ce^(-x) into the above equation, we get:
2x - Ce^(-x) - x^2 - Ce^(-x) = 2x - x^2
Simplifying and canceling terms, we get:
Ce^(-x) = x^2
Therefore, C = x^2*e^(x) and substituting this value in equation (1), we get:
y = x^2 + xe^(-x)
This is the particular solution of the given differential equation.
Now, let's check the other given differential equations for exactness:
y' + xy = x^3 + 2x:
This equation is not exact since M_y = 1 and N_x = 0. To find the integrating factor, we can use the formula:
IF = e^(∫x dx) = e^(x^2/2)
Multiplying both sides of the equation by this integrating factor, we get:
e^(x^2/2)y' + xe^(x^2/2)y = x^3e^(x^2/2) + 2xe^(x^2/2)
The left-hand side of the equation is now exact, so we can find a potential function f(x,y) such that df/dx = e^(x^2/2)y and df/dy = xe^(x^2/2). Integrating df/dx, we get:
f(x,y) = ∫e^(x^2/2)y dx = (1/2)e^(x^2/2)y + g(y)
Differentiating f(x,y) with respect to y and equating it to xe^(x^2/2), we get:
(1/2)e^(x^2/2) + g'(y) = xe^(x^2/2)
Solving for g(y), we get:
g(y) = 0
Substituting this value in the expression for f(x,y), we get:
f(x,y) = (1/2)e^(x^2/2)y
Therefore, the general solution to the differential equation is given by:
(1/2)e^(x^2/2)y = ∫(x^3 + 2x)e^(x^2/2) dx = (1/2)e^(x^2/2)(x^2 + 1) + C,
where C is a constant. Rearranging, we get:
y = (x^2 + 1) + Ce^(-x^2/2)
x*y' + y = 3x^2:
This equation is exact since M_y = 1 and N_x = 1. We can find the potential function f(x,y) such that df/dx = x and df/dy = 1 by integrating both sides of the given equation with respect to x and y, respectively. We get:
f(x,y) = (1/2)x^2 + ln|y| + g(y)
Taking the partial derivative with respect to y and equating it to 1, we get:
(1/y) + g'(y) = 1
Solving for g(y), we get:
g(y) = ln|y| + C
Substituting this value in the expression for f(x,y), we get:
f(x,y) = (1/2)x^2 + ln|y| + C
Therefore, the general solution to the differential equation is given by:
(1/2)x^2 + ln|y| = C
Substituting the initial condition y(0) = 1 into the above equation, we get:
C = (1/2)(0)^2 + ln|1| = 0
Therefore, the particular solution to the differential equation with the initial condition y(0) = 1 is:
(1/2)x^2 + ln|y| = 0
ln|y| = -(1/2)x^2
|y| = e^(-(1/2)x^2)
y = ±e^(-(1/2)x^2)
Learn more about equation from
https://brainly.com/question/29174899
#SPJ11
A manufacturing company produces two models of an HDTV per week, x units of model A and y units of model B with a cost (in dollars) given by the following function.
C(x,y)=3x^2+6y^2
If it is necessary (because of shipping considerations) that x+y=90, how many of each type of set should be manufactured per week to minimize cost? What is the minimum cost? To minimize cost, the company should produce units of model A. To minimize cost, the company should produce units of model B. The minimum cost is $
The answer is 15 and 75 for the number of model A and model B sets produced per week, respectively.
Given: C(x, y) = 3x² + 6y²x + y = 90
To find: How many of each type of set should be manufactured per week to minimize cost? What is the minimum cost?Now, Let's use the Lagrange multiplier method.
Let f(x,y) = 3x² + 6y²
and g(x,y) = x + y - 90
The Lagrange function L(x, y, λ)
= f(x,y) + λg(x,y)
is: L(x, y, λ)
= 3x² + 6y² + λ(x + y - 90)
The first-order conditions for finding the critical points of L(x, y, λ) are:
Lx = 6x + λ = 0Ly
= 12y + λ = 0Lλ
= x + y - 90 = 0
Solving the above three equations, we get: x = 15y = 75
Putting these values in Lλ = x + y - 90 = 0, we get λ = -9
Putting these values of x, y and λ in L(x, y, λ)
= 3x² + 6y² + λ(x + y - 90), we get: L(x, y, λ)
= 3(15²) + 6(75²) + (-9)(15 + 75 - 90)L(x, y, λ)
= 168,750The minimum cost of the HDTVs is $168,750.
To minimize the cost, the company should manufacture 15 units of model A and 75 units of model B per week.
To know more about number visit:
https://brainly.com/question/3589540
#SPJ11
7. Form the differential equation by eliminating the orbitary constant from \( y^{2}=4 a x \). 8. Solve \( y d x+x d y=e^{-x y} d x \) if cuts the \( y \)-axis.
7. The required differential equation is [tex]\[y \frac{d y}{d x}=2 a\][/tex]
8. The solution of the given differential equation if it cuts the y-axis is [tex]\[y=y_{0}{{e}^{xy}}\].[/tex]
7. Differential equation : [tex]\[y^{2}=4 a x\][/tex]
To eliminate the arbitrary constant [tex]\[a\][/tex], take [tex]\[\frac{d}{d x}\][/tex] on both sides and simplify.
[tex]\[\frac{d}{d x}\left( y^{2} \right)=\frac{d}{d x}\left( 4 a x \right)\]\[2 y \frac{d y}{d x}=4 a\]\[y \frac{d y}{d x}=2 a\][/tex]
Therefore, the required differential equation is [tex]\[y \frac{d y}{d x}=2 a\][/tex]
8. Given differential equation: [tex]\[y d x+x d y=e^{-x y} d x\][/tex]
We need to find the solution of the given differential equation if it cuts the y-axis.
Since the given differential equation has two variables, we can not solve it directly. We need to use some techniques to solve this type of differential equation.
If we divide the given differential equation by[tex]\[d x\][/tex], then it becomes \[tex][y+\frac{d y}{d x}e^{-x y}=0\][/tex]
We can write this in a more suitable form as [tex][\frac{d y}{d x}+\left( -y \right){{e}^{-xy}}=0\][/tex]
This is a linear differential equation of the first order. The general solution of this differential equation is given by
[tex]\[y={{e}^{\int{(-1{{e}^{-xy}}}d x)}}\left( \int{0{{e}^{-xy}}}d x+C \right)\][/tex]
This simplifies to
[tex]\[y=C{{e}^{xy}}\][/tex]
Now we need to find the value of the constant [tex]\[C\][/tex].
Since the given differential equation cuts the y-axis, at that point the value of [tex]\[x\][/tex] is zero. Therefore, we can substitute [tex]\[x=0\][/tex] and [tex]\[y=y_{0}\][/tex] in the general solution to find the value of [tex]\[C\][/tex].[tex]\[y_{0}=C{{e}^{0}}=C\][/tex]
Therefore, [tex]\[C=y_{0}\][/tex]
Hence, the solution of the given differential equation if it cuts the y-axis is [tex]\[y=y_{0}{{e}^{xy}}\][/tex].
Learn more about differential equations:
https://brainly.com/question/9686230
#SPJ11
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.
Learn more about induction proof:
https://brainly.com/question/30401663
#SPJ11
Verify that y = (c1+c2t)e^t + sin(t) +t² is a solution to y"-2y'+y=-2 cos(t) +t²-4t+2,
where C1, C2 ∈R are arbitrary constants.
By substitution and simplification, we have shown that [tex]\(y = (c_1 + c_2t)e^t + \sin(t) + t^2\)[/tex]is indeed a solution to the given differential equation.
To verify that [tex]\(y = (c_1 + c_2t)e^t + \sin(t) + t^2\)[/tex] is a solution to the given differential equation, we need to substitute this expression for \(y\) into the equation and check if it satisfies the equation.
Let's start by finding the first and second derivatives of \(y\) with respect to \(t\):
[tex]\[y' = (c_2 + c_2t + c_1 + c_2t)e^t + \cos(t) + 2t,\]\[y'' = (2c_2 + c_2t + c_2 + c_2t + c_1 + c_2t)e^t - \sin(t) + 2.\][/tex]
Now, substitute these derivatives into the differential equation:
[tex]\[y'' - 2y' + y = (2c_2 + c_2t + c_2 + c_2t + c_1 + c_2t)e^t - \sin(t) + 2 - 2((c_2 + c_2t + c_1 + c_2t)e^t + \cos(t) + 2t) + (c_1 + c_2t)e^t + \sin(t) + t^2.\][/tex]
Simplifying this expression, we get:
[tex]\[2c_2e^t + 2c_2te^t + 2c_2e^t - 2(c_2e^t + c_2te^t + c_1e^t + c_2te^t) + c_1e^t + c_2te^t - \cos(t) + 2 - \cos(t) - 4t + 2 + (c_1 + c_2t)e^t + \sin(t) + t^2.\][/tex]
Combining like terms, we have:
[tex]\[2c_2e^t + 2c_2te^t - 2c_2e^t - 2c_2te^t - 2c_1e^t - \cos(t) + 2 - \cos(t) - 4t + 2 + c_1e^t + c_2te^t + \sin(t) + t^2.\][/tex]
Canceling out terms, we obtain:
\[-2c_1e^t - 4t + 4 + t^2 - 2\cos(t).\]
This expression is equal to \(-2\cos(t) + t^2 - 4t + 2\), which is the right-hand side of the given differential equation.
Learn more about differential equation here :-
https://brainly.com/question/32645495
#SPJ11
Let X⊆R^d be a set of d+1 affinely independent points. Show that int(conv(X))=∅.
a) √(1/3)a³. √12a² : √2a b) √(27x³y^{5}) : √(1/3)xy
d) 3x.(√27x^{5} : √(1/3)x³)
We have proved that if X ⊆ R^d is a set of d+1 affinely independent points, then int(conv(X)) ≠ ∅.
Given that X ⊆ R^d is a set of d+1 affinely independent points, we need to prove that int(conv(X)) ≠ ∅.
Definition: A set of points in Euclidean space is said to be affinely independent if no point in the set can be represented as an affine combination of the remaining points in the set.
Solution:
In order to show that int(conv(X)) ≠ ∅, we need to prove that the interior of the convex hull of the given set X is not an empty set. That is, there must exist a point that is interior to the convex hull of X.
Let X = {x_1, x_2, ..., x_{d+1}} be the set of d+1 affinely independent points in R^d. The convex hull of X is defined as the set of all convex combinations of the points in X. Hence, the convex hull of X is given by:
conv(X) = {t_1 x_1 + t_2 x_2 + ... + t_{d+1} x_{d+1} | t_1, t_2, ..., t_{d+1} ≥ 0 and t_1 + t_2 + ... + t_{d+1} = 1}
Now, let us consider the vector v = (1, 1, ..., 1) ∈ R^{d+1}. Note that the sum of the components of v is (d+1), which is equal to the number of points in X. Hence, we can write v as a convex combination of the points in X as follows:
v = (d+1)/∑i=1^{d+1} t_i (x_i)
where t_i = 1/(d+1) for all i ∈ {1, 2, ..., d+1}.
Note that t_i > 0 for all i and t_1 + t_2 + ... + t_{d+1} = 1, which satisfies the definition of a convex combination. Also, we have ∑i=1^{d+1} t_i = 1, which implies that v is in the convex hull of X. Hence, v ∈ conv(X).
Now, let us show that v is an interior point of conv(X). For this, we need to find an ε > 0 such that the ε-ball around v is completely contained in conv(X). Let ε = 1/(d+1). Then, for any point u in the ε-ball around v, we have:
|t_i - 1/(d+1)| ≤ ε for all i ∈ {1, 2, ..., d+1}
Hence, we have t_i ≥ ε > 0 for all i ∈ {1, 2, ..., d+1}. Also, we have:
∑i=1^{d+1} t_i = 1 + (d+1)(-1/(d+1)) = 0
which implies that the point u = ∑i=1^{d+1} t_i x_i is a convex combination of the points in X. Hence, u ∈ conv(X).
Therefore, the ε-ball around v is completely contained in conv(X), which implies that v is an interior point of conv(X). Hence, int(conv(X)) ≠ ∅.
Learn more about independent points here :-
https://brainly.com/question/31987907
#SPJ11
Find the equation of the tangent line to the graph of f(x) = √x+81 at the point (0,9).
Answer:
dy/dx = 1/2 x ^(-1/2)
gradient for point (0,9) = 1/6
y-0 = 1/6 (x-9)
y = 1/6 (x-9)
What is the solution to the system of equations in the graph below?
The solution to the system of equations is x = -8 and y = -28.
To find the solution to the system of equations, we can use either the substitution method or the elimination method. Let's use the elimination method for this example.
Step 1: Multiply the second equation by 2 to make the coefficients of y in both equations equal:
2(x - 2y) = 2(48)
2x - 4y = 96
Now, we have the following system of equations:
2x - y = 12
2x - 4y = 96
Step 2: Subtract the first equation from the second equation to eliminate the variable x:
(2x - 4y) - (2x - y) = 96 - 12
2x - 4y - 2x + y = 84
-3y = 84
Step 3: Solve for y by dividing both sides of the equation by -3:
-3y / -3 = 84 / -3
y = -28
Step 4: Substitute the value of y back into one of the original equations to solve for x. Let's use the first equation:
2x - (-28) = 12
2x + 28 = 12
2x = 12 - 28
2x = -16
x = -8
So, the solution to the system of equations 2x - y = 12 and x - 2y = 48 is x = -8 and y = -28.
To know more about equation here
https://brainly.com/question/21835898
#SPJ4