All categories

3 years ago

Let g(x) = 2x and h(x) = x2 + 4. Evaluate (h ∘ g)(−2).

Mathematics

2 answers:

kap26 [50]3 years ago

7 0

Let g(x) = 2x and h(x) = x2 + 4. Evaluate (h ∘ g)(−2).

A. −12

the corrects answer is <em>B. −16 </em>

C. 20

D. 16

Nastasia [14]3 years ago

7 0

Answer:

Option C, 20 is the answer.

Step-by-step explanation:

Let g(x) = 2x and h(x) = x² + 4

We have to evaluate (h ο g) ( -2 )

Since ( h ο g )x = h [ g(x) ]

= ( 2x)² + 4

= 4x² + 4

( h ο g ) ( -2 ) = 4 ( -2 )² + 4

= 4 × 4 + 4

= 16 + 4

= 20

Option C, 20 will be the answer.

You might be interested in

Select the correct answer from each drop-down menu.

ella [17]

Answer:

A non-equilateral rhombus.

Step-by-step explanation:

We can solve this graphically.

We start with square:

ABCD

with:

A = (11, - 7)

B = (9, - 4)

C = (11, - 1)

D = (13, - 4)

Only with the vertices, we can see that ABCD is equilateral, as the length of each side is:

AB = √( (11 - 9)^2 + (-7 -(-4))^2) = √( (2)^2 + (3)^2) = √(4 + 9) = √13

BC = √( (11 - 9)^2 + (-1 -(-4))^2) = √13

CD = √( (11 - 13)^2 + (-1 -(-4))^2) = √13

DA = √( (11 - 13)^2 + (-7 -(-4))^2) = √13

And we change C by C' = (11, 1)

In the image you can see the 5 points and the figure that they make:

The figure ABCD is a rhombus, and ABC'D is also a rhombus, the only difference between the figures is that ABCD is equilateral while ABC'D is not equilateral.

5 0

2 years ago

A battery has a lifetime that is approximately normally distributed with a mean of 600 hours and a standard deviation of 50 hour

miss Akunina [59]

Answer:

A. Your battery is likely defective since such poor performance is extremely unlikely.

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If X is more than two standard deviations from the mean, it is considered an unlikely outcome.

In this question, we have that:

$\mu = 600, \sigma = 50$