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ryzh [129]
3 years ago
8

If the parent function f(x) = (2x − 3)3 is transformed to g(x) = (-2x + 3)3, which type of transformation occurs? A. vertical sh

ift B. horizontal shift C. horizontal reflection D. vertical reflection

Mathematics
2 answers:
tigry1 [53]3 years ago
5 0

Answer:

c plato answer

Step-by-step explanation:

exis [7]3 years ago
4 0

Answer:

The answer is C

Step-by-step explanation:

We have f(x) function as below:

f(x)=(2x-3)^3

and the converted function g(x) is as mentioned below:

g(x)=(-2x+3)^3

We can observe that:

f(x)=-g(x)

Therefore it can be said that this function reflected horizontally.

Note: please check the plot attached.

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You can represent the measures of an angle and its complement as x° and (90 − x)°. Similarly, you can represent the measures of
UNO [17]
X = 4(180 - x)
x = 720 - 4x
4x + x = 720
5x = 720
x = 720/5
x = 144

180 - x =
180 - 144 =
36

the measure of the angle is 144...the measure of its supplement is 36
4 0
3 years ago
A car travels 3 miles. It's tires make 2640 revolutions. What is the radius of a tire in inches?
Alexus [3.1K]

Answer:

7929

Step-by-step explanation:

it is 7929 because if u times 3 and 2640 it's 7929

5 0
3 years ago
The table shows the price for different numbers of board games:
NikAS [45]

Answer

B

Step-by-step explanation:

Multiply all the Number of  Board Games by 3.5. It is the same as the Price already showed on the graph. So, the correct answer is B.

3 0
3 years ago
A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and
pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean \mu and standard deviation \sigma, we have these following probabilities

Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827

Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545

Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So \mu = 53, \sigma = 11

So:

Pr(53-11 \leq X \leq 53+11) = 0.6827

Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545

Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973

-----------

Pr(42 \leq X \leq 64) = 0.6827

Pr(31 \leq X \leq 75) = 0.9545

Pr(20 \leq X \leq 86) = 0.9973

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have Pr(31 \leq X \leq 75) = 0.9545 = 0.9545 subtracted by Pr(42 \leq X \leq 64) = 0.6827 is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.

To find just the percentage above the mean, we divide this value by 2

So:

P = {0.9545 - 0.6827}{2} = 0.1359

The approximate percentage of cars that remain in service between 64 and 75 months is 13.59%.

4 0
3 years ago
Jamie ordered 24 combo meals for $5 each for a party. The service charge for home delivery for the whole purchase was $6.
Agata [3.3K]
It's easy you do 24 times 5 and then find out the service charge by dividing ratios between the meal and parties and then add 6$ to your total!
7 0
3 years ago
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