Is the function linear or quadratic?

Mathematics

1 answer:

fgiga [73]3 years ago

7 0

It is quadratic since a quadratic function is of the form :y=ax²+bx+c ,a,b,c are constants.

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If f(x)=x^2-7 and h(x)=7x-5, then f(h(2))=

zaharov [31]

We are given 3 equations. Using these equations we can find f(h(2)).

h(x)=7x-5
h(2)=7(2)-5
h(2)=14-5
h(2)=9

f(x)=x^2-7
f(h(2))=(7x-5)^2-7
we know the answer for h(2) by now, so we can substitute for that.
f(9)=9^2-7
f(9)=81-7
f(9)=74

so f(h(2))=74

5 0

3 years ago

Which statement best explains the relationship between

-Dominant- [34]

Answer: They are not perpendicular because their slopes are not negative reciprocals.

Step-by-step explanation:

For these two lines to be perpendicular, their slopes have to be negative reciprocals.

Slope of FG.

F = (-4, 1)

G = (0, -2)

= (Y₂ - Y₁) / (X₂ - X₁)

= (-2 - 1) / (0 + 4)

= -3/4

Slope of HJ

J = (0 , 4)

H = (-4, -2)

Slope = (Y₂ - Y₁) / (X₂ - X₁)

= (-2 - 4) / (-4 -0)

= 6/4

= 3/2

<em>The slopes of these lines are not negative reciprocals so these lines are not perpendicular. </em>

4 0

3 years ago

2x^2+5x+3/x^2-3x-4 divided by 4x^2+2x-6/x^2-8x+16

Masja [62]

Answer: $\frac{x-4}{2x-2}$

Step-by-step explanation:

$\frac{2x^2+5x+3}{x^2-3x-4}$ divided by $\frac{4x^2+2x-6}{x^2-8x+16}$ is the same thing as multiplying $\frac{2x^2+5x+3}{x^2-3x-4}$ by $\frac{x^2-8x+16}{4x^2+2x-6}$ .