Ask question

Ask question

All categories

4 years ago

9

An equation is shown below:

1 answer:

maw [93]4 years ago

3 0

Part A: 1 solution

Part B:
4(3d −7) = −52

(4)(3d) + (4)(-7)= -52

12d + -28 = -52

12d - 28 = -52

12d - 28 + 28 = -52 + 28

12d = -24

12d/12 = -24/12

d = -2

:)

You might be interested in

100 points if correct In this task, you will play the role of a restaurant owner. You will collect data and analyze different as

jek_recluse [69]

Answer:

Step-by-step explanation:

7 0

3 years ago

Oliver earns five dollars for each lawn he mowed.If he he’d six lawns to move but forgot to mow three of them how much money did

snow_lady [41]

$15

Explanation:

For each lawn he mows = $5

<u>He should mow 6 lawns but did only 3 lawns as other 3 he forgot to mow.</u>

earned:

mows lawn-d * $5

3 * $5

$15

3 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

eduard

The Lagrangian is

$L(x_1,\ldots,x_n,\lambda_1,\ldots,\lambda_n)=x_1+\cdots+x_n+\lambda_1({x_1}^2+\cdots+{x_n}^2)+\cdots+\lambda_n({x_1}^2+\cdots+{x_n}^2)$

with partial derivatives (set equal to 0)

$\dfrac{\partial L}{\partial x_i}=1+2x_i(\lambda_1+\cdots+\lambda_n)=0$

$\dfrac{\partial L}{\partial\lambda_i}={x_1}^2+\cdots+{x_n}^2-36=0$

for each $1\le i\le n$ .

Let $\Lambda$ be the sum of all the multipliers $\lambda_i$ ,

$\Lambda=\displaystyle\sum_{k=1}^n\lambda_k=\lambda_1+\cdots+\lambda_n$

We notice that

$x_i\dfrac{\partial L}{\partial x_i}=x_i+2{x_i}^2\Lambda=0$

so that

$\displaystyle\sum_{i=1}^nx_i\dfrac{\partial L}{\partial x_i}=\sum_{i=1}^nx_i+2\Lambda\sum_{i=1}^n{x_i}^2=0$

We know that $\sum\limits_{i=1}^n{x_i}^2=36$ , so

$\displaystyle\sum_{i=1}^nx_i+2\Lambda\sum_{i=1}^n{x_i}^2=0\implies\sum_{i=1}^nx_i=-72\Lambda$

Solving the first $n$ equations for $x_i$ gives

$1+2\Lambda x_i=0\implies x_i=-\dfrac1{2\Lambda}$

and in particular

$\displaystyle\sum_{i=1}^nx_i=-\dfrac n{2\Lambda}$

It follows that

$-\dfrac n{2\Lambda}+72\Lambda=0\implies\Lambda^2=\dfrac n{144}\implies\Lambda=\pm\dfrac{\sqrt n}{12}$

which gives us

$x_i=-\dfrac1{2\left(\pm\frac{\sqrt n}{12}\right)}=\pm\dfrac6{\sqrt n}$

That is, we've found two critical points,

$\pm\left(\dfrac6{\sqrt n},\ldots,\dfrac6{\sqrt n}\right)$

At the critical point with positive signs, $f(x_1,\ldots,x_n)$ attains a maximum value of

$\displaystyle\sum_{i=1}^nx_i=\dfrac{6n}{\sqrt n}=6\sqrt n$

and at the other, a minimum value of

$\displaystyle\sum_{i=1}^nx_i=-\dfrac{6n}{\sqrt n}=-6\sqrt n$

4 0

4 years ago

NEED HELP! WILL MARK BRAINLIEST

ICE Princess25 [194]

Answer:

im pretty sure its 27.

Step-by-step explanation:

if g(x) = 20 then the new problem is 20=x-7

20=x-7 . add 7 to both sides

+7=+7 . +7 cancels out the -7

27=x . <<<answer

6 0

3 years ago

Olivia has $5.50 dimes and quarters. She has 8 more quarters than dimes. How many quarters does she have

Radda [10]

18 quarters
just do trial and error. It's easy!

8 0

4 years ago