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3 years ago

Which function has an inverse that is a function?

Mathematics

2 answers:

tresset_1 [31]3 years ago

8 0

Answer:

All but the second one

Step-by-step explanation:

Find the inverse of each function:

b(x)=x^2+3

y=x^2+3

x=y^2+3

x-3=y^2

sqrt(x-3)=y

This inverse function is a function

d(x)=-9

y=-9

This doesn't have an inverse function

m(x)=-7x

y=-7x

x=-7y

-1/7x=y

This inverse function is a function

p(x)=|x|

y=|x|

x=|y|

x=y

This inverse function is a function

Therefore, all but the second function have inverses that are functions

Jobisdone [24]3 years ago

6 0

Answer:

All but the second one have inverse functions

Step-by-step explanation:

Brainliest pls pls pls pls

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Given the definitions of f(x) and g(x) below, find the value of g(f(-1)).

Reptile [31]

Answer:

-14

Step-by-step explanation:

$f(x) = -4x – 7 \\ plug \: x = - 1 \\ f( - 1) = - 4( - 1) - 7 = 4 - 7 = - 3 \\g(x) = x² + 7x – 2 \\g(f( - 1)) = {( - 3)}^{2} + 7( - 3) - 2 \\ = 9 - 21 - 2 \\= - 14$

5 0

3 years ago

What’s the answer pleazzzzz helplllllll 9m^2-4m-6 by 3m

dlinn [17]

Answer

D. 3m - 4/3 -2/m

<u>Explanation</u>

(9m²-4m-6) ÷ 3m = 9m²/3m - 4m/3m - 6/3m

Carry out the division, each term is divided by 3m.

9m²/3m = 3m

- 4m/3m = -4/3

- 6/3m = - 2/m

Finally you will get,

5 0

3 years ago

Helppppppp please:( and thank you

Shtirlitz [24]

Answer:

180-140=50 180-90-50= 40 m=40

3 0

3 years ago

A student S is suspected of cheating on exam, due to evidence E of cheating being present. Suppose that in the case of a cheatin

Kay [80]

Answer:

Step-by-step explanation:

Suppose we think of an alphabet X to be the Event of the evidence.

Also, if Y be the Event of cheating; &

Y' be the Event of not involved in cheating

From the given information:

$P(\dfrac{X}{Y}) = 60\% = 0.6$

$P(\dfrac{X}{Y'}) = 0.01\% = 0.0001$

$P(Y) = 0.01$

Thus, $P(Y') \ will\ be = 1 - P(Y)$

P(Y') = 1 - 0.01

P(Y') = 0.99

The probability of cheating & the evidence is present is = P(YX)

$P(YX) = P(\dfrac{X}{Y}) \ P(Y)$

$P(YX) =0.6 \times 0.01$

$P(YX) =0.006$

The probabilities of not involved in cheating & the evidence are present is:

$P(Y'X) = P(Y') \times P(\dfrac{X}{Y'})$

$P(Y'X) = 0.99 \times 0.0001 \\ \\ P(Y'X) = 0.000099$

(b)

The required probability that the evidence is present is:

P(YX or Y'X) = 0.006 + 0.000099

P(YX or Y'X) = 0.006099

(c)

The required probability that (S) cheat provided the evidence being present is:

Using Bayes Theorem

$P(\dfrac{Y}{X}) = \dfrac{P(YX)}{P(Y)}$

$P(\dfrac{Y}{X}) = \dfrac{P(0.006)}{P(0.006099)}$

$P(\dfrac{Y}{X}) = 0.9838$

5 0

3 years ago

Lot the points (0, −2) and (4, 1) on the coordinate plane.

S_A_V [24]

$(0,\ -2)\to x=0,\ y=-2$

from orgin (0, 0):

0 units left-right

2 units down

$(4,\ 1)\to x=4,\ y=1$

4 units right

1 unit up

6 0

4 years ago