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Ber [7]
3 years ago
5

Please help! functions operations. explain please

Mathematics
2 answers:
Ronch [10]3 years ago
8 0
<h2>Hello!</h2>

The answer is: a. 40

<h2>Why?</h2>

We are working with composite functions, so, to find the correct option, first we need to follow the composite function process:

(h\circ g)=h(g(x))

So,

We are given the functions:

g(x)=2x\\\\h(x)=x^{2}+4

So,

(h\circ g)(x)=(2x)^{2}+4

Then, evaluating with "x" equal to -3, we have:

(h\circ g)(-3)=(2*-3)^{2}+4=(-6)^{2}+4=36+4=40

So, the correct option is a. 40

Have a nice day!

ValentinkaMS [17]3 years ago
7 0

Answer:

a. 40

Step-by-step explanation:

Given

g(x)=2x and h(x)=x^2+4

(h\circ g)(x)=h(g(x))

(h\circ g)(x)=h(2x)

We plug in 2x into  h(x)=x^2+4.

(h\circ g)(x)=(2x)^2+4

(h\circ g)(x)=4x^2+4

We now substitute x=-3.

(h\circ g)(-3)=4(-3)^2+4

(h\circ g)(-3)=36+4=40

The correct choice is A.

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What value of x makes this equation true? 6x/-4=6(-1/2)<br> a) -2<br> b)2<br> c)3<br> d)-3
Yuri [45]
6x/-4 = 6(-1/2)
6x/-4 = - 3...multiply by -4
6x = -3 * -4
6x = 12
x = 2


7 0
3 years ago
5x-9=15+11x<br><br> The solution for x is x=?
Wittaler [7]

Answer:

x = -4

Step-by-step explanation:

   5x-9 = 15+11x

5x - 11x = 15 + 9

      -6x = 24

         x = -4

5 0
2 years ago
Read 2 more answers
The variable y varies jointly with x and w when y = - 42, x = 2, and w = -3.
Andrei [34K]

Answer:

k = 7

Step-by-step explanation:

given y varies jointly with x and w then the equation relating them is

y = kxw ← k is the constant of variation

to find k use the condition when y = - 42, x = 2 and w = - 3 , then

- 42 = k × 2 × - 3

- 42 = - 6k ( divide both sides by - 6 )

7 = k

constant of variation is k = 7

6 0
1 year ago
PLEASE HELP!! FINALS
denis-greek [22]

Answer:   22.8


Step-by-step explanation:


6 0
3 years ago
From a population of 200 elements, the standard deviation is known to be 14. a sample of 49 elements is selected. it is determin
Savatey [412]
N = population size
n = sample size
sigma = population standard deviation
xbar = sample mean
SE = standard error
fpc = finite population correction

In this case,
N = 200
n = 49
sigma = 14
xbar = 56

Since n/N = 49/200 = 0.245 is larger than 0.05, this means we must use a finite population correction factor. I'll use fpc in place of 'finite population correction'.
If we ignore the fpc, then the SE would be simply sigma/sqrt(n) = 14/sqrt(49) = 2.
However we cannot ignore the fpc. We must use it due to the fact that n/N > 0.05. 

--------------------------

Let's compute the fpc factor
fpc = sqrt((N-n)/(N-1))
fpc = sqrt((200-49)/(200-1))
fpc = 0.87108780834612

--------------------------

With the fpc factor, we'll have the true SE to be SE = fpc*sigma/sqrt(n) = 0.87108780834612*14/sqrt(49) = 1.74217561669224

The final answer, accurate to 6 decimal places, is therefore 1.742176
6 0
3 years ago
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