Evaluate 3+jk+k^3 when j=2 and k=6

Question 3 (2 points) Saved Which of the following has the larger area? A rectangle with sides 12 inches and 15 inches. A triang

ZanzabumX [31]

Answer:

rectangle = 180

triangle = 350

the triangle has a much larger area.

350-180=170

the triangle by 17

square inches

5 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

andrey2020 [161]

Answer:

The maximum value is 1/27 and the minimum value is 0.

Step-by-step explanation:

Note that the given function is equal to $(xyz)^2$ then it means that it is positive i.e $f(x,y,z)\geq 0$ .

Consider the function $F(x,y,z,\lambda)=x^2y^2z^2-\lambda (x^2+y^2+z^2-1)$

We want that the gradient of this function is equal to zero. That is (the calculations in between are omitted)

$\frac{\partial F}{\partial x} = 2x(y^2z^2 - \lambda)=0$

$\frac{\partial F}{\partial y} = 2y(x^2z^2 - \lambda)=0$

$\frac{\partial F}{\partial z} = 2z(x^2y^2 - \lambda)=0$

$\frac{\partial F}{\partial \lambda} = (x^2+y^2+z^2-1)=0$

Note that the last equation is our restriction. The restriction guarantees us that at least one of the variables is non-zero. We've got 3 options, either 1, 2 or none of them are zero.

If any of them is zero, we have that the value of the original function is 0. We just need to check that there exists a value for lambda.

Suppose that x is zero. Then, from the second and third equation we have that

$-2y\lambda = -2z\lambda$ . If lambda is not zero, then y =z. But, since $-2y\lambda=0$ and lambda is not zero, this implies that x=y=z=0 which is not possible. This proofs that if one of the variables is 0, then lambda is zero. So, having one or two variables equal to zero are feasible solutions for the problem.

Suppose that only x is zero, then we have the solution set $y^2+z^2=1$ .

If both x,y are zero, then we have the solution set $z^2=1$ . We can find the different solution sets by choosing the variables that are set to zero.

NOw, suppose that none of the variables are zero.

From the first and second equation we have that

$\lambda = y^2z^2 = x^2z^2$ which implies $x^2=y^2$

Also, from the first and third equation we have that

$\lambda = y^2z^2 = x^2y^2$ which implies $x^2=z^2$

So, in this case, replacing this in the restriction we have $3z^2=1$ , which gives as another solution set. On this set, we have $x^2=y^2=z^2=\frac{1}{3}$ . Over this solution set, we have that the value of our function is $\frac{1}{3^3}= \frac{1}{27}$

4 0

3 years ago

Stan wants to start an IRA that will have $250,000 in it when he retires in 25 years. How much should he invest semiannually in

tester [92]

Answer:

10000

Step-by-step explanation:

5 0

3 years ago

I'm not sure what they mean by "relations that represents a function". I don't understand this.

sergeinik [125]

Means, for every 1 input, there is 1 coresponding output
normally x is input and y is output in form (x,y)
so just look for the option(s) that has/have every first number repeat with only 1 2nd number

A. -7 repeats with 5 and -3, not a function

B. no repeats of first number (6,-8) and (-5,-8) are fine because first numbers don't repeat, das is funciton

C. no repeats, function

D. no repeats, function

answer is B,C,D

4 0

3 years ago

Solve x 2/3> 8 or 2/3x < 4.

siniylev [52]

Answer:

2.56 repeating

Step-by-step explanation:

8 0

3 years ago