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Black_prince [1.1K]
3 years ago
13

Can someone show me how to do this ?

Mathematics
1 answer:
Savatey [412]3 years ago
7 0
The point slope formula is y - y1 = m(x - x1)

the value x1 is the x coordinate of a point
the value y1 is the y coordinate of a point
the value m is the slope, which is 3

therefore, x1 = -2 & y1 = 1

Then, all you do is plug in the points:

y - 1 = 3(x + 2)

NOTE:
It is 3(x + 2) instead of 3(x - 2) because the formula is (x - x1). If you plug x1 in, it is (x - (-2)). 2 negatives cancel out to be a positive, so it is (x + 2)

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Given ∆ABC, with a = 9 , b = 8 and c =13, find angle c
strojnjashka [21]

Answer:

angle c is 13 as it said in the question

Step-by-step explanation:

6 0
2 years ago
A stack of one hundred fifty cards is placed next to a ruler, and the height of the stack is measured to be 5/8 inches.
DaniilM [7]
Divide 5/8 by 150, to see how thick each is, that'll give you 150 even pieces that add up to 5/8

\bf \cfrac{\frac{5}{8}}{150}\implies \cfrac{\frac{5}{8}}{\frac{150}{1}}\implies \cfrac{5}{8}\cdot \cfrac{1}{150}\implies \cfrac{1}{8\cdot 30}\implies \cfrac{1}{240}
5 0
3 years ago
IT IS DUE AT 11:59PM. PLEASE HELP!!!!
kaheart [24]

Hello from MrBillDoesMath!

Answer:   b = 4, h = 13

Discussion:

For a triangle with base b" and height "h, the area formula is

A=  (1/2)bh.

In our case A = 26, h = 5b-7, so

26 = (1/2) b (5b-7).  Multiply both sides by 2:

26*2 = b(5b-7) =>

52 = b(5b-7)  =>

52 = 5b^2 - 7b =>           (subtract 52 from both sides)

5b^2 - 7b - 52 = 0.   This factors as follows:

(b - 4) (5 b + 13) = 0        (use quadratic formula to find this)

so b = 4 and h = 5b -7 = 5*4 - 7 = 13


Thank you,

MrB


3 0
2 years ago
Read 2 more answers
Given that f(x) = 2x + 5 and g(x) = x − 7, solve for f(g(x)) when x = −3
dusya [7]

Answer:

-15

Step-by-step explanation:

start from the inside and go out.

So first plug in -3 into g(x)

g(-3) = -3 - 7 = -10

then plug in -10 into f(x)

f(-10) = 2(-10) + 5 = -15

so f(g(x)) = -15

5 0
3 years ago
Read 2 more answers
Does (0,0) satisfy x+3y>3
PtichkaEL [24]

\bf x+3y > 3\qquad \qquad (\stackrel{x}{0}~~,~~\stackrel{y}{0})\qquad \qquad 0+3(0) > 3\implies \stackrel{\stackrel{\textit{is this true?}}{\downarrow }}{0>3}\qquad nope!

4 0
3 years ago
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