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JulsSmile [24]
3 years ago
13

PLEASE HELP BEFORE 10:00AM EST PLEASE!!!!!!!!! PLEASE SHOW WORK IF YOU CAN!!

Mathematics
1 answer:
Tju [1.3M]3 years ago
3 0
1. answer is $246.68 (rounded) because 230 x 1.0725 (multiplier) is $246.675.
2. answer is B
3. answer is A rounded because you will get (51.5962...)
4. answer is A because when divide 40 by 42.59 you get 1.06475 (but that is the multiplier) so the answer is A
5. answer is A because 12000x 1.04 is 12480 and 12480-12000=480
6.I guess the answer is D but the total he earns is $2440 but from the one car he earns $2240 +$200 (which he earns anyway)
7.answer is A $9.16
8. the answer is B
(J-)Hope this helps
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a rectangle swimming pool is 4 ft deep. one side of the pool is 2.5 times longer than the other. the amount of water needed to f
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Find the are of the pool by dividing the volume by the depth:

1960 / 4 = 490

The pool is 490 square feet.


Area is found by multiplying the length by the width.

Let the width = X

We are told the length is 2.5X ( 2.5 times longer)


So we now have 2.5x * x = 490


2.5x * x = 2.5x^2


Now we have 2.5x^2 = 490


Divide both sides by 2.5:

x^2 = 490/2.5

x^2 = 196

find X by taking the square root of 196:

x = √196

x = 14


The width is 14 feet

The length is 2.5 * 14 = 35 feet

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Pleas help me if this is correct
Vesna [10]

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Step-by-step explanation:

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2 years ago
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Find the midpoint of the line segment with endpoints at the given coordinates (-6,6) and (-3,-9)
JulsSmile [24]

The midpoint of the line segment with endpoints at the given coordinates (-6,6) and (-3,-9) is \left(\frac{-9}{2}, \frac{-3}{2}\right)

<u>Solution:</u>

Given, two points are (-6, 6) and (-3, -9)

We have to find the midpoint of the segment formed by the given points.

The midpoint of a segment formed by \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { and }\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right) is given by:

\text { Mid point } \mathrm{m}=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)

\text { Here in our problem, } x_{1}=-6, y_{1}=6, x_{2}=-3 \text { and } y_{2}=-9

Plugging in the values in formula, we get,

\begin{array}{l}{m=\left(\frac{-6+(-3)}{2}, \frac{6+(-9)}{2}\right)=\left(\frac{-6-3}{2}, \frac{6-9}{2}\right)} \\\\ {=\left(\frac{-9}{2}, \frac{-3}{2}\right)}\end{array}

Hence, the midpoint of the segment is \left(\frac{-9}{2}, \frac{-3}{2}\right)

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3 years ago
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