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2 years ago

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What are slope and y intercept of the graph of the linear function shown above ?

1 answer:

Mashutka [201]2 years ago

3 0

Answer:

d. slope= -2 and y-intercept= 5

Step-by-step explanation:

i hope this helps :)

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I need a answer to this

Ugo [173]

-6b = -66

You divide -6 on both sides. A negative multiplied/divided by a negative equals a positive

-1 · -1 = 1

-1 ÷ -1 = 1

b = 11

7 0

3 years ago

What is the y-intercept of the equation of the line that is perpendicular to the line y =

Luda [366]

Answer:

$y$ intercept $=10$

line $y=-x+10$

Step-by-step explanation:

Let $y=mx+b$ is the required line.

<u>Slope of perpendicular lines :</u> If $m_{1}$ and $m_{2}$ be slope of two perpendicular lines. Them $m_{1}\times m_{2}=-1$

$y=mx+b$ is perpendicular to $y=x+10$

the slope of $y=x+10$ is $=1$

$m\times 1=-1\\m=-1$

Hence eqn is $y=-x+b$

This equation passes through $(15,-5)$

$-5=-15+b\\b=-5+15\\b=10$

so the line is $y=-x+10$

<u>y-intercept :</u>

Substitute $x=10$

$y=0+10\\y=10$

7 0

3 years ago

HELP also if you don't know how to do it than don't add an answer it make me super mad.

dlinn [17]

H(17) = 17
Do you need help with the other 2?

4 0

2 years ago

Find an expression for a cubic function f if f(4) = 96 and f(-4) = f(0) = f(5) = 0. Part 1 of 4 A cubic function generally has t

Bezzdna [24]

Given a polynomial $p(x)$ and a point $x_0$ , we have that

$p(x_0) = 0 \iff (x-x_0) \text{\ divides\ } p(x)$

We know that our cubic function is zero at -4, 0 and 5, which means that our polynomial is a multiple of

$(x+4)(x)(x-5) = x(x+4)(x-5)$

Since this is already a cubic polynomial (it's the product of 3 polynomials with degree one), we can only adjust a multiplicative factor: our function must be

$f(x) = ax(x+4)(x-5),\quad a \in \mathbb{R}$

To fix the correct value for a, we impose $f(4)=96$ :

$f(4) = 4a(4+4)(4-5) = -32a = 96$

And so we must impose

$-32a=96 \iff a = -\dfrac{96}{32} = -3$

So, the function we're looking for is

$f(x) = -3x(x+4)(x-5)=-3x^3+3x^2+60x$

4 0

2 years ago

Solve the following integral.<br><br> <img src="https://tex.z-dn.net/?f=%5Cint4x%5Ccos%282-3x%29dx" id="TexFormula1" title="\int

bagirrra123 [75]

Hi there!

$\boxed{-\frac{4x}{3}sin(2-3x) + \frac{4}{9}cos(2-3x) + C}$

To find the indefinite integral, we must integrate by parts.

Let "u" be the expression most easily differentiated, and "dv" the remaining expression. Take the derivative of "u" and the integral of "dv":

u = 4x

du = 4

dv = cos(2 - 3x)

v = 1/3sin(2 - 3x)

Write into the format:

∫udv = uv - ∫vdu

Thus, utilize the solved for expressions above:

4x · (-1/3sin(2 - 3x)) -∫ 4(1/3sin(2 - 3x))dx

Simplify:

-4x/3 sin(2 - 3x) - ∫ 4/3sin(2 - 3x)dx

Integrate the integral:

∫4/3(sin(2 - 3x)dx

u = 2 - 3x

du = -3dx ⇒ -1/3du = dx

-1/3∫ 4/3(sin(2 - 3x)dx ⇒ -4/9cos(2 - 3x) + C

Combine:

$-\frac{4x}{3}sin(2-3x) + \frac{4}{9}cos(2-3x) + C$

7 0

2 years ago

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