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gtnhenbr [62]
3 years ago
7

What is the slope of a line that passes through (–12,15) and (0,4)?

Mathematics
2 answers:
Dmitrij [34]3 years ago
6 0
Slope for points (x1,y1) and (x2,y2)
slope=(y2-y1)/(x2-x1)

(x,y)
(-12,15)
(0,4)

x1=-12
y1=15
x2=0
y2=4

slope=(4-15)/(0-(-12))=-11/(0+12)=-11/12


slope=-11/12
Romashka [77]3 years ago
5 0
I think that the answer is negative 11 over 12
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Given:

The equation is

\dfrac{???}{x-2}/\dfrac{x^2-1}{x^2-4x+4}=\dfrac{5(x-2)}{x-1}

To find:

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Solution:

Let the missing value be k.

\dfrac{k}{x-2}/\dfrac{x^2-1}{x^2-4x+4}=\dfrac{5(x-2)}{x-1}

\dfrac{k}{x-2}\times \dfrac{x^2-2(x)(2)+2^2}{x^2-1^2}=\dfrac{5(x-2)}{x-1}

Using the formulae (a-b)^2=a^2-2ab+b^2 and (a-b)(a+b)=a^2-b^2.

\dfrac{k}{x-2}\times \dfrac{(x-2)^2}{(x-1)(x+1)}=\dfrac{5(x-2)}{x-1}

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Cancel out the common factors from both sides.

\dfrac{k}{x+1}=5

Multiply both sides by (x+1).

k=5(x+1)

Therefore, the missing value is 5(x+1).

5 0
4 years ago
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