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3 years ago

17% of 200 books is ....

Mathematics

2 answers:

nadya68 [22]3 years ago

8 0

34 books would be the answer.

VikaD [51]3 years ago

5 0

34 is 17% of 200 if that is what your asking

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What is the algebraic representation (rule) of a figure that has been dilated using a scale factor of 2?

Illusion [34]

Answer:

Okay 2

Step-by-step explanation:

Okay since that is that you gotta that 23u172, got it?

5 0

2 years ago

3(2x+1)=21 <br> Solve this multi-step equation using the distributive property

Tasya [4]

3 (2x + 1) = 21
6x + 3= 21
-3 -3
6x = 18

Divide each side by 6, and x equals 3.

3 (2(3) + 1) = 21
3 (6 + 1) = 21
18 + 3 = 21

Hope this helps! Can I please have brainliest so I can level up? Thank you!

7 0

2 years ago

Find an ordered pair for x and y

Yuliya22 [10]

Answer: here are a few ordered pairs

1. (0,-4)

2.(8,0)

3.(16,4)

7 0

3 years ago

HELP ME PLEASE, I'M GIVING YOU ALL MY POINTS ;-; (which is 25)

pishuonlain [190]

The answer to 31,096 + 59,721 is 89,817 so B 89,800 is the best estimate

2nd- The answer to 69,896 - 42,901 is 26,995 so A 26,900 is the best answer

3rd- 630,429 - 27,811 is 602,618 so the best answer(s) would be A 630,400 and C 27,800

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