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3 years ago

8

A line segment has endpoints at (-7, 2) and (8,8). What is the x-coordinate of the point on the line segment that is the distanc

e from (-7, 2) to (8,8)?

2 answers:

slava [35]3 years ago

8 0

Answer:

Sorry dud it been long time I've not answered 0ne

sveticcg [70]3 years ago

6 0

Answer:

Heyy

Step-by-step explanation:

Did ya get the answer ???

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Maru [420]

Answer:

x=32

Step-by-step explanation:

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2 years ago

Log2(y)=3log3(u)<br> log5(y)=2log(u) + 3log5(v)<br> Write y in terms of u and v

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2 years ago

A store pays $10 for a blanket, and the markup is 115%. A customer will also pay a 5 1/2% sales tax. What will be the total cost

Naddik [55]

Firstly, we need to find the price of the blanket/bracelet after the 115% increase.
$10 x 1.15 = $15.20

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$15.20 x 0.055 = 0.83
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3 years ago

Solve x^2 – 7x = -13.

lara31 [8.8K]

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2 years ago

Read 2 more answers

The length of each side of an equilateral triangle is increased by 20%, resulting in triangle ABC. If the length of each side of

kompoz [17]

Answer: Area of ΔABC is 2.25x the area of ΔDEF.

Step-by-step explanation: Because equilateral triangle has 3 equal sides, area is calculated as

$A=\frac{\sqrt{3} }{4} a^{2}$

with a as side of the triangle.

Triangle ABC is 20% bigger than the original, which means its side (a₁) measures, compared to the original:

a₁ = 1.2a

Then, its area is

$A_{1}=\frac{\sqrt{3} }{4}(1.2a)^{2}$

$A_{1}=\frac{\sqrt{3} }{4}1.44a^{2}$

Triangle DEF is 20% smaller than the original, which means its side is:

a₂ = 0.8a

So, area is

$A_{2}=\frac{\sqrt{3} }{4} (0.8a)^{2}$

$A_{2}=\frac{\sqrt{3} }{4} 0.64a^{2}$

Now, comparing areas:

$\frac{A_{1}}{A_{2}}= (\frac{\sqrt{3}.1.44a^{2} }{4})(\frac{4}{\sqrt{3}.0.64a^{2} } )$

$\frac{A_{1}}{A_{2}} =$ 2.25

<u>The area of ΔABC is </u><u>2.25x</u><u> greater than the area of ΔDEF.</u>

7 0

2 years ago