Solve the system by elimination:
36x-63y=7
-24x+42y=0

Answers

Answer 1
Answer:

Answer:sorry I don't think itis possible with elimination method if it is possible you can post it in the comment section

Step-by-step explanation:


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Place parentheses in the expression so that it equals the given value 12 = 2 X 23; value: 120

Answers

No the value answer is 48

A box is formed by cutting squares from the four corners of a sheet of paper and folding up the sides.Suppose the paper is 7"-wide by 9"-long.
a. Estimate the maximum volume for this box?
b. What cutout length produces the maximum volume?

Answers

To answer this question it is necessary to find the volume of the box as a function of "x", and apply the concepts of a maximum of a function.

The solution is:

a) V (max) = 36.6 in³

b) x = 1.3 in

The volume of a cube is:

V(c) = w×L×h  ( in³)

In this case, cutting the length  "x" from each side, means:

wide of the box    ( w - 2×x )   equal to  ( 7 - 2×x )

Length of the box ( L - 2×x )   equal to  ( 9 - 2×x )

The height  is  x

Then the volume of the box,  as a function of x is:

V(x) = ( 7 - 2×x ) × ( 9 -2×x ) × x

V(x) = ( 63 - 14×x - 18×x + 4×x²)×x

V(x) = 4×x³ - 32×x² + 63×x

Tacking derivatives,  on both sides of the equation

V´(x) = 12×x² - 64 ×x + 63

If   V´(x) = 0      then      12×x² - 64 ×x + 63 = 0

This expression is a second-degree equation, solving for x

x₁,₂ = [ 64 ± √ (64)² - 4×12*63

x₁ =  ( 64 + 32.74 )/ 24

x₁ = 4.03     this value  will bring us an unfeasible solution,  since it is not possible to cut 2×4 in from a piece of paper of 7 in ( therefore we dismiss that value)

x₂ = ( 64 - 32.74)/24

x₂ = 1.30 in

The  maximum volume of the box is:

V(max) = ( 7 - 2.60) × ( 9 - 2.60)×1.3

V(max) = 4.4 × 6.4 × 1.3

V(max) = 36.60 in³

To chek for maximum value of V when x = 1.3

we find the second derivative of V  V´´,  and substitute the value of x = 1.3,    if the relation is smaller than 0,  we have a maximum value of V

V´´(x) = 24×x - 64 for x = 1.3

V´´(x) = 24× 1.3 - 64            ⇒   V´´(x) < 0

Then the value  x = 1.3 will bring maximum value for V

Related Link: brainly.com/question/13581879

Final answer:

The maximum volume of the box that can be formed is approximately 17.1875 cubic inches. The cutout length that would result in this maximum volume is approximately 1.25 inches.

Explanation:

To solve this problem, we will use optimization in calculus. Let's denote the length of the square cutout as 'x'. When you cut out an x by x square from each corner and fold up the sides, the box will have dimensions:

  • Length: 9 inches (the original length) - 2x (the removed parts)
  • Width: 7 inches (the original width) - 2x
  • Height: x inches (the height is the cutout's length)

So the volume V of the box can be given by the equation: V = x(9-2x)(7-2x). We want to maximize this volume.

To find the maximum, differentiate V with respect to x, equate to zero and solve for x. V' = (9-2x)(7-2x) + x(-2)(7-2x) + x(9-2x)(-2) = 0. We obtain x=1.25 inches, but we need to verify whether this value gives us a maximum. Second differentiation, V'' = -12 is less than zero for these dimensions so the V is maximum.

a. So, when we solve, the maximum volume will be approximately 17.1875 cubic inches.

b. The cutout length that would produce the maximum volume is therefore about 1.25 inches.

Learn more about Optimization here:

brainly.com/question/37742146

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-3 < 5 true or false

Answers

Answer:

true

Step-by-step explanation:

Answer:

true?

Step-by-step explanation:

Find the size of angle x.

Answers

Answer:

111°

Step-by-step explanation:

By exterior angle theorem:

x = 82 \degree + 29 \degree \n  = 111 \degree

Answer:

x = 111

Step-by-step explanation:

180- 82 -29 is 69 and 180-69 is x which x is equal to 111

What is the solution to the equation One-fourth x minus one-eighth = Start Fraction 7 Over 8 End Fraction + one-half x? x = negative 5 x = negative 4 x = 4 x = 5.

Answers

Answer:x = negative 4

Step-by-step explanation:

The given equation is expressed as

1/4 × x - 1/8 = 7/8 + 1/2 × x

x/4 - 1/8 = 7/8 + x/2

First step is to find the lowest common multiple of the left hand side of the equation and the right hand side of the equation. Then, we would multiply both sides of the equation by the lowest common multiple. The lowest common multiple is 8. Therefore

x/4 × 8 - 1/8 × 8 = 7/8 × 8 + x/2 × 8

2x - 1 = 7 + 4x

7 + 4x = 2x - 1

Subtracting 2x and 7 from the left hand side of the equation and the right hand side of the equation, it becomes

7 - 7 + 4x - 2x = 2x - 2x - 1 - 7

2x = - 8

x = - 8/2 = - 4

Answer:

Step-by-step explanation:

-4

g let X be a normally distributed random variable with mean 3 and variance 4. a) Let Y = 5X+2. What is the distribution of Y? What are its mean and variance? b) Find P(Y<10). Find P(X<10). c) What is the 99th percentile of the distribution of Y? d) What is the 99th percentile of the distribution of X? e) What is the distribution of W = exp(Y)? What are its mean and variance?

Answers

a. Let F_X(x) be the CDF of X. The CDF of Y is

F_Y(y)=P(Y\le y)=P(5X+2\le y)=P\left(X\le\frac{y-2}5\right)=F_X\left(\frac{y-2}5\right)

which is to say, Y is also normally distributed, but with different parameters. In particular,

E[Y]=E[5X+2]=5E[X]+2=17

\mathrm{Var}[Y]=\mathrm{Var}[5X+2]=5^2\mathrm{Var}[X]=100

b. Using the appropriate CDFs, we have

P(Y<10)=F_Y(10)=F_X\left(\frac{10-2}5\right)=F_X(1.6)\approx0.242

P(X<10)=F_X(10)\approx0.9998

c. The 99th percentile for any distribution D is the value of d_(0.99) such that P(D\le d_(0.99))=0.99, i.e. all values of d below d_(0.99) make up the lower 99% of the distribution.

We have

P(Y\le y_(0.99))=0.99\implies y_(0.99)\approx40.26

d. On the other hand, the 99th percentile for X is

P(X\le x_(0.99))=0.99\implies x_(0.99)\approx7.653

e. We have

F_W(w)=P(W\le w)=P\left(e^Y\le w\right)=P(Y\le\ln w)=F_Y(\ln w)

which suggests that \ln W is normally distributed, or W is log-normally distributed. Recall that the moment-generating function for Y is

M_Y(t)=\exp\left(17t+\frac{100t^2}2\right)

But we also have

M_Y(t)=E[e^(tY)]=E[e^(t\ln W)]=E[W^t]

Then

E[W]=M_Y(1)=e^(67)

and

E[W^2]=M_Y(2)=e^(234)\implies\mathrm{Var}[W]=E[W^2]-E[W]^2=e^(234)-e^(134)