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3 years ago

How can you use functions to model linear relationships?

Mathematics

1 answer:

Oliga [24]3 years ago

6 0

Answer:

a function is just a comparison between two quantities so when you go to plot the points, the input is x and the output is y.

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which of the following plotted points on the graph represent the zeros of the function g(x) = (x^2 - 3x - 10)(x + 4)

Natalka [10]

Answer: The answer would be (-2,0),(-4,0), and (5,0)

Step-by-step explanation:

Guessing the answer choices are

(-2,0)

(4,0)

(0,10)

(-4,0)

(0,-2)

(0,-10)

(5,0)

3 0

2 years ago

FP!!!!!!!!!!!!!!!!!!! if ur up ( 10 x 8 ) + 150

taurus [48]

Answer:

220

Step-by-step explanation:

first we have by Bracket of BODMAS rule then we have to multiply 10 *8 = 80

first we have by Bracket of BODMAS rule then we have to multiply 10 *8 = 80then we have to open bracket 80 +150

first we have by Bracket of BODMAS rule then we have to multiply 10 *8 = 80then we have to open bracket 80 +150 = 220

(10*8) +15

Solution

Solution= (80) +150

Solution= (80) +150= 80+150

Solution= (80) +150= 80+150=. 220

Solution= (80) +150= 80+150=. 220

8 0

2 years ago

Read 2 more answers

Simplify . 4y4 4y8 8y4 8y8

Tanya [424]

Answer:

Step-by-step explanation:

Just did it.

4 0

3 years ago

If al|b, mZ1 = 10x + 2 and mZ4 = 5x + 13 , find mZ1 and m<2

morpeh [17]

Um un um i’m i’m in i’m i’m um um u me m um i’m in ...21

5 0

3 years ago

Suppose r(t)=costi+sintj+3tk represents the position of a particle on a helix, where z is the height of the particle above the g

Ilia_Sergeevich [38]

a. The $\vec k$ component tells you the particle's height:

$3t=16\implies t=\dfrac{16}3$

b. The particle's velocity is obtained by differentiating its position function:

$\vec v(t)=\dfrac{\mathrm d\vec r(t)}{\mathrm dt}=-\sin t\,\vec\imath+\cos t\,\vec\jmath+3\,\vec k$

so that its velocity at time $t=\frac{16}3$ is

$\vec v\left(\dfrac{16}3\right)=-\sin\dfrac{16}3\,\vec\imath+\cos\dfrac{16}3\,\vec\jmath+3\,\vec k$

c. The tangent to $\vec r(t)$ at $t=\frac{16}3$ is

$\vec T(t)=\vec r\left(\dfrac{16}3\right)+\vec v\left(\dfrac{16}3\right)t$

4 0

3 years ago