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3 years ago

Which of the following functions is not a linear function?

Mathematics

1 answer:

Bad White [126]3 years ago

5 0

Answer:

Hello,

not linear function :

f(x) = -2

f(x) = x²

f(x) = x - 2

Step-by-step explanation:

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[Inequalities] Please help and thanks.

kondor19780726 [428]

Answer:

First blank is >=

Second blank is <=

Step-by-step explanation:

At least means the sign is greater than or equal to

No more than means the sign is less than or equal to

5 0

3 years ago

Determine whether the sets have a subset relationship. R = {right triangles} O = {obtuse triangles} a) R and O are disjoint. b)

DerKrebs [107]

Answer:

Step-by-step explanation:

A right angled triangle has one angle equal to 90°

an obtuse angle triangle has one angle greater tan 90°

a) An o and right-angled triangle btuse angled triangle can have one angle equal.

b) R and O are not equivalent by definition

c) subset of R and Oare not all triangles as R and O are categories of all triangles. R and O are subset of all triangles

d) Acute triangles have all acute angles. So subset of R and O can't be all acute traingles

e) All triangles with two acute angles are may or may not have third angle as obtuse angle or 90° angle

f) none of the above

5 0

3 years ago

If the parent function f(x) = x^2 is fatter translated 11 units to the left, then translated 5 units down, write the resulting f

nikklg [1K]

The quadratic function given by:

$f(x)=a(x-h)^2+k, \ \ \ a\neq 0$

is in vertex form. The graph of $f$ is a parabola whose axis is the vertical line $x=h$ and whose vertex is the point $(h, k)$ . So:

To translate the graph of a function to the right, left, upward or downward we have:

$For \ a \ positive \ real \ number \ c. \ \mathbf{Vertical \ and \ horizontal \ shifts} \\ in \ the \ graph \ of \ y=f(x) \ are \ represented \ as \ follows:\\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{upward}: \\ g(x)=f(x)+c \\ \\ \bullet \ Vertical \ shift \ c \ units \ \mathbf{downward}: \\ g(x)=f(x)-c \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ \mathbf{right}: \\ g(x)=f(x-c) \\ \\ \bullet \ Horizontal \ shift \ c \ units \ to \ the \ \mathbf{left}: \\ g(x)=f(x+c)$

By knowing this things, we can solve our problem as follows:

FIRST.

Translating 11 units to the left:

$g(x)=f(x+11) \\ \\ \therefore g(x)=(x+11)^2$

Then translating 5 units down:

$g(x)=f(x)-c \\ \\ \therefore g(x)=(x+11)^2-5$

Since the new function is fatter, the factor we need to multiply the term $(x+11)^2$ must be less than 1, to make the graph fatter. So, according to our options, there are two factors 1/2 and 2.