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2 years ago

10

Rise over run algebra one help please

1 answer:

Ainat [17]2 years ago

4 0

Answer:

Slope is -3

Equation is y = -3x + 2

Step-by-step explanation:

(0 , 2) and ( -1 , 5)

m= (y2-y1)/(x2-x1)

m= (5 - 2)/(-1 -0)

m= 3/-1

m= -3

y - y1 = m(x - x1)

y - 2 = -3(x - 0)

y - 2 = -3x + 0

y = -3x + 2

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7. f(x) = Square root of quantity x plus nine. ; g(x) = 8x - 13 Find f(g(x)). (1 point) f(g(x)) = 2 Square root of quantity two

AlladinOne [14]

Answer:

f(g(x)) =

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Step-by-step explanation:

f(x) =

$\sqrt{x} \: + \: 9$

g(x) = 8x - 13

f(g(x) =

$\sqrt{8x \: - 13} \: + \: 9$

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What does c equal in this equation 6c + 11 = 2c + 59

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2 years ago

Use the equation and type the ordered-pairs. y = log 3 x {(1/3, a0), (1, a1), (3, a2), (9, a3), (27, a4), (81, a5)

vagabundo [1.1K]

Answer:

Considering the given equation $y = log_{3}x\\$

And the ordered pairs in the format $(x, y)$

I don't know if it is log of base 3 or 10, but I will assume it is 3.

For $(\frac{1}{3}, a_{0} )$

$x=\frac{1}{3}$

$y=a_{0}$

$y = log_{3}x\\y = log_{3}(\frac{1}{3} )\\y=-\log _3\left(3\right)\\y=-1$

So the ordered pair will be $(\frac{1}{3}, -1 )$

For $(1, a_{1} )$

$x=1$

$y=a_{1}$

$y = log_{3}x\\y = log_{3}1\\y = log_{3}(1)\\Note: \log _a(1)=0\\y = 0$

So the ordered pair will be $(1, 0 )$

For $(3, a_{2} )$

$x=3$

$y=a_{2}$

$y = log_{3}x\\y = log_{3}3\\y = 1$

So the ordered pair will be $(3, 1 )$

For $(9, a_{3} )$

$x=9$

$y=a_{3}$

$y = log_{3}x\\y = log_{3}9\\y=2\log _3\left(3\right)\\y=2$

So the ordered pair will be $(9, 2 )$

For $(27, a_{4} )$

$x=27$

$y=a_{4}$

$y = log_{3}x\\y = log_{3}27\\y=3\log _3\left(3\right)\\y=3$

So the ordered pair will be $(27, 3 )$

For $(81, a_{5} )$

$x=81$

$y=a_{5}$

$y = log_{3}x\\y = log_{3}81\\y=4\log _3\left(3\right)\\y=4$

So the ordered pair will be $(81, 4 )$

4 0

3 years ago

-6-12:(+3) <br> Efectueaza

sleet_krkn [62]

The answer is -10.
I hope this is right...

4 0

2 years ago

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