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2 years ago

11

The length of a rectangle is 5 more than twice the width. Which equation could represent the area of the rectangle in terms of t

he width?

2 answers:

Daniel [21]2 years ago

6 0

A=lw

w=width

2w+5=length

42=w(2w+5)

42=2w2+5w

2w2+5w-42=0

use the quadratic equation

(-5+√25+336)/4

(-5+√361)/4

(-5+19)/4

14/4=3.5

w=3.5 m

2w+5=12 m

3.5 by 12 rectangle

liq [111]2 years ago

5 0

Answer:

l=5+2w

Step-by-step explanation:

Since the length is twice the width plus 5, the equation would look like :

l=5+2w

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577 rounded to the nearest ten

anygoal [31]

The correct answer 580

Hope this helps

7 0

3 years ago

Read 2 more answers

That propertand of an arraand is still in the equation k (A + B) = kA + kB

Anarel [89]

Answer:

that propertand of an arraand is in the equation

select the appropriate response

c Transposition

6 0

3 years ago

Evaluate the surface integral:S

rjkz [21]

Assuming $S$ does not include the plane $z=0$ , we can parameterize the region in spherical coordinates using

$\mathbf r(u,v)=\left\langle3\cos u\sin v,3\sin u\sin v,3\cos v\right\rangle$

where $0\le u\le2\pi$ and $0\le v\le\dfrac\pi/2$ . We then have

$x^2+y^2=9\cos^2u\sin^2v+9\sin^2u\sin^2v=9\sin^2v$
$(x^2+y^2)=9\sin^2v(3\cos v)=27\sin^2v\cos v$

Then the surface integral is equivalent to

$\displaystyle\iint_S(x^2+y^2)z\,\mathrm dS=27\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^2v\cos v\left\|\frac{\partial\mathbf r(u,v)}{\partial u}\times \frac{\partial\mathbf r(u,v)}{\partial u}\right\|\,\mathrm dv\,\mathrm du$

We have

$\dfrac{\partial\mathbf r(u,v)}{\partial u}=\langle-3\sin u\sin v,3\cos u\sin v,0\rangle$
$\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle3\cos u\cos v,3\sin u\cos v,-3\sin v\rangle$
$\implies\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}=\langle-9\cos u\sin^2v,-9\sin u\sin^2v,-9\cos v\sin v\rangle$
$\implies\left\|\dfrac{\partial\mathbf r(u,v)}{\partial u}\times\dfrac{\partial\mathbf r(u,v)}{\partial v}\|=9\sin v$

So the surface integral is equivalent to

$\displaystyle243\int_{u=0}^{u=2\pi}\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv\,\mathrm du$
$=\displaystyle486\pi\int_{v=0}^{v=\pi/2}\sin^3v\cos v\,\mathrm dv$
$=\displaystyle486\pi\int_{w=0}^{w=1}w^3\,\mathrm dw$

where $w=\sin v\implies\mathrm dw=\cos v\,\mathrm dv$ .

$=\dfrac{243}2\pi w^4\bigg|_{w=0}^{w=1}$
$=\dfrac{243}2\pi$

4 0

3 years ago

The following condensed information was reported by Peabody Toys, Inc., for 2021 and 2020: ($ in thousands) 2021 2020 Income sta

never [62]

Answer:

The answers are given below.

Step-by-step explanation:

The computation is shown below:

1.a.

Profit Margin = Net Income ÷ Sales × 100

= $374 ÷ $6,900 ×100

= 5.4%

1-b:

Average Assets = (Beginning Assets + Ending Assets) ÷ 2

= ($3,200 + $3,600) ÷ 2

= $3,400

Now

Return on Assets = Net Income ÷ Average Assets

= $374 ÷ $3,400

= 11%

1-c

Average Equity = ($700 + $700 + $320 + $270) ÷ 2

= $995

Now

Return on Equity = Net Income ÷ Average Equity *100

= $374 ÷ $995

= 37.59%

2:

Dividends Paid = Beginning Retained Earnings + Net Income – Ending Retained Earnings

= $270 + $374 - $320

= $324

4 0

3 years ago

An animal feed to be mixed from soybean meal and oats must contain at least 168 lb of protein, 27 lb of fat, and 14 lb of minera

atroni [7]

Answer:

The animal farm should buy 1.775 bags of soybeans and 1.575 bags of oats

$Cost = 51.275$

Step-by-step explanation:

The given parameters can be summarized as:

$\begin{array}{cccc}{} & {x} & {y} & {Total} & {Protein} & {70} & {21} & {168} &{Fats}& {9} & {7} & {27} & {Minerals} & {7} & {1} & {14}& {Cost} & {21} & {7} & {} \ \end{array}$

Where: x = Soybeans and y = Oats.

So, the system of equations are:

$70x + 21y = 168$

$9x +7y = 27$

$7x + y = 14$

$Cost = 21x + 7y$

The best way to solve this, is using graph

Plot the following equations on a graph, and get the points of intersection:

$70x + 21y = 168$

$9x +7y = 27$

$7x + y = 14$

From the attached graph, we have:

$(x_1,y_1) = (1.636,2.545)$

$(x_2,y_2) = (1.775,1.575)$

$(x_3,y_3) = (2.023,1.256)$

Substitute each of the values of x's and y's in the cost function to get the minimum cost:

$Cost = 21x + 7y$

$(x_1,y_1) = (1.636,2.545)$

$Cost = 21 * 1.636 + 7 * 2.545$

$Cost = 52.171$

$(x_2,y_2) = (1.775,1.575)$

$Cost = 21 * 1.775 + 7 * 1.575$

$Cost = 48.3$

$(x_3,y_3) = (2.023,1.256)$

$Cost = 21 * 2.023 + 7 * 1.256$

$Cost = 51.275$

The values of x and y that gives the minimum cost is:

$(x_2,y_2) = (1.775,1.575)$

and the minimum cost is:

$Cost = 48.3$

Hence, the animal farm should buy 1.775 bags of soybeans and 1.575 bags of oats

5 0

3 years ago