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3 years ago

9

Create a set of coordinates to model a relation that is a linear function.

1 answer:

STALIN [3.7K]3 years ago

3 0

I will create a set of arbitrary constants (x1,y1) (x2,y2)

slope = y2-y1/x2-x1

y = (y2-y1/x2-x1)x + b

y2 = (y2-y1/x2-x1)x2 + b

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

y = (y2-y1/x2-x1)x + [y2 - (y2-y1/x2-x1)]

Choose any points and just
Plug the values and you have a linear function.

NOT SURE IF THAT'S WHAT THE QUESTION WANTS.

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3 years ago

Identify each function that has a remainder of -3 when divided x+6

Sergeu [11.5K]

Answer:

D

Step-by-step explanation:

According to remainder theorem, you can know the remainder of these polynomials if you plug in x = -6 into them.

<em>So we will plug in -6 into x of all the polynomials ( A through D) and see which one equals -3.</em>

<em />

<em>For A:</em>

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For B:

$x^4 + 4x^3 - 21x^2 - 53x + 12\\=(-6)^4 + 4(-6)^3 - 21(-6)^2 - 53(-6) + 12\\=6$

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For D:

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2 years ago

Write the equation for the hyperbola with foci (–12, 6), (6, 6) and vertices (–10, 6), (4, 6).

fomenos

Answer:

$\frac{(x--3)^2}{49} -\frac{(y-6)^2}{32}=1$

Step-by-step explanation:

The standard equation of a horizontal hyperbola with center (h,k) is

$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1$

The given hyperbola has vertices at (–10, 6) and (4, 6).

The length of its major axis is $2a=|4--10|$ .

$\implies 2a=|14|$

$\implies 2a=14$

$\implies a=7$

The center is the midpoint of the vertices (–10, 6) and (4, 6).

The center is $(\frac{-10+4}{2},\frac{6+6}{2}=(-3,6)$

We need to use the relation $a^2+b^2=c^2$ to find $b^2$ .

The c-value is the distance from the center (-3,6) to one of the foci (6,6)

$c=|6--3|=9$

$\implies 7^2+b^2=9^2$

$\implies b^2=9^2-7^2$

$\implies b^2=81-49$

$\implies b^2=32$

We substitute these values into the standard equation of the hyperbola to obtain:

$\frac{(x--3)^2}{7^2} - \frac{(y-6)^2}{32}=1$

$\frac{(x+3)^2}{49} -\frac{(y-6)^2}{32}=1$

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