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3 years ago

8

GIVING BRAINLIEST

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Delicious77 [7]3 years ago

3 0

Sarah was locked out of her apartment and the only way out is through a window three stories high 20 feet above the ground. She needed to borrow from one of her friends. There is a tree along the side of the house so she will need to put the ladder 15 feet from her apartment. How long does the ladder need to be to reach the apartment window?

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Susie gains more weight over Christmas than any other time of the year. Christmas makes Susie gain weight. This conclusion is...

Dahasolnce [82]

Hello there!

Your answer is D) not justifiable, because there is another variable to consider.
While Susie may gain weight during Christmas, the holiday is not magically making her gain weight. It is more likely all of those candy canes.

I hope I helped!

Let me know if you need anything else!

~ Zoe

3 0

3 years ago

Read 2 more answers

HELP ME ASAP!!!! PLEASE HELP I AM DESPERATE!!!!!!!

evablogger [386]

First question
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Second question 65%

5 0

4 years ago

a farmer is planning a rectangular area for her chickens. the area of the rectangle will be 200 square feet. three sides of the

Vladimir79 [104]

Let

x---------> the length side of the rectangular area

y---------> the width side of the rectangular area

we know that

the area of the rectangle is equal to

$A=x*y\\ A= 200\ ft^{2} \\ x*y=200$

$y=\frac{200}{x}$ -----> equation $1$

The perimeter of the rectangle is equal to

$P=2x+2y$

but remember that the fourth side of the rectangle will be formed by a portion of the barn wall

so

$P=x+2y$ -----> equation $2$

<em>To minimize the cost we must minimize the perimeter</em>

Substitute the equation $1$ in the equation $2$

$P=x+2*[\frac{200}{x} ]$

Using a graph tool

see the attached figure

The minimum of the graph is the point $(20,40)$

that means for $x=20\ ft$

the perimeter is a minimum and equal to $40\ ft$

<u>Find the value of y</u>

$y=\frac{200}{x}$

$y=\frac{200}{20}$

$y=10\ ft$

The cost of fencing is equal to

$5*40= \$200$

therefore

<u>the answer is</u>

the length side of the the fourth wall will be $20\ ft$

7 0

3 years ago

A rectangle has a perimeter of 34 cm in an area of 52 cm squared it's length is 5 more than twice its width write and solve the

valina [46]

Let $l$ be the length and $w$ be the width of the rectangle.

Given perimeter of the rectangle

$2(l+w)=34 \; l+w=17$

Area of the rectangle is

$lw=52$

Also length is 5 more than twice its width,

$l=2w+5$

From the above equations

$(l-w)^2=(l+w)^2-4lw=17^2-4*52=81\\ l-w=9,\; since \; l>w$

Now $2l=9+17=26\;,l=13,w=4$

The dimensions of the rectangle are length $l=13$ , and width $w=4$ .

8 0

4 years ago

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Can pls someone help I need help

Firlakuza [10]

A^2 + b^2 = c^2
9^2 + b^2 = 15^2
81 + b^2 = 225
225 - 81 = b^2
144 = b^2
square root of 144 = 12
b = 12

7 0

3 years ago