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Alekssandra [29.7K]
3 years ago
9

Write an equation in slope-intercept form for the line that satisfies the following condition. slope 6, and passes through (5, 2

0)
Mathematics
2 answers:
evablogger [386]3 years ago
4 0
I can't remember if this is correct or not but I believe it would be 5 multiplied by 20 plus 6. As I said I don't Know for sure if this is correct or not. However, I hope it help
EastWind [94]3 years ago
4 0
First you graph..


Start at (5,20)


Count 6 up 1 to the right three or four times then Count 6 down and 1 to the left.

You will get the y-intercept out of this...

Which is -10.

You can put it into slope-intercept form now-

y=mx+b

"m" being slope and "b" being the y-intercept.

Final Answer: y=6x+10
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3 years ago
A population of 40 rats living near a restaurant dumpster is growing at a monthly rate of 50%
goldenfox [79]

[(40 ÷ .5) Y] + 40

Parentheses are rat growth

Brackets are rat growth by how many years or just exclude that and the brackets if you don't need to know how many years are there are

The 40 is the amount of rats already there

Growth of rats plus the amount already there.

I'm probably wrong lol

7 0
3 years ago
Let f(x)=11-3xm compute f(2)+f^-1(2)
Ganezh [65]
Taking f(X) to be y
Y = 11 - 3X
3X = 11 - y
X  = (11-y)÷3
F^-1(x) = (11 - x)÷3 addING both equations
It gives = 0

Please mark as brainliest
8 0
3 years ago
Simplify. Thank you...
Verizon [17]

Answer:

I'm not entirely sure but I got:

\frac{ {4x}^{3} + 3 }{(2x - 1)( {x}^{2}  + x)( {2x}^{2} - 3x + 1) }

5 0
3 years ago
Read 2 more answers
Enter the correct answer in the box. solve the equation x2 − 16x 54 = 0 by completing the square. fill in the values of a and b
poizon [28]

The roots of the given polynomials exist  $x=8+\sqrt{10}$, and $x=8-\sqrt{10}$.

<h3>What is the formula of the quadratic equation?</h3>

For a quadratic equation of the form $a x^{2}+b x+c=0$ the solutions are

$x_{1,2}=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Therefore by using the formula we have

$x^{2}-16 x+54=0$$

Let, a = 1, b = -16 and c = 54

Substitute the values in the above equation, and we get

$x_{1,2}=\frac{-(-16) \pm \sqrt{(-16)^{2}-4 \cdot 1 \cdot 54}}{2 \cdot 1}$$

simplifying the equation, we get

$&x_{1,2}=\frac{-(-16) \pm 2 \sqrt{10}}{2 \cdot 1} \\

$&x_{1}=\frac{-(-16)+2 \sqrt{10}}{2 \cdot 1}, x_{2}=\frac{-(-16)-2 \sqrt{10}}{2 \cdot 1} \\

$&x=8+\sqrt{10}, x=8-\sqrt{10}

Therefore, the roots of the given polynomials are $x=8+\sqrt{10}$, and

$x=8-\sqrt{10}$.

To learn more about quadratic equations refer to:

brainly.com/question/1214333

#SPJ4

3 0
1 year ago
Read 2 more answers
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