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LUCKY_DIMON [66]

3 years ago

11

Identify the maximum and minimum values of the function y = 8 cos x in the interval [-2π, 2π]. Use your understanding of transfo

rmations, not your graphing calculator.

1 answer:

Lina20 [59]3 years ago

5 0

Pls. see attachment for the answer.

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It takes 8 minutes for 3 cars to move through a car wash. At the same rate, how many cars can move through the car wash in 24 mi

Marizza181 [45]

9 cara is the answer

4 0

4 years ago

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Subtract 1/6 a+3 from 1/3 a−5 .

steposvetlana [31]

$\frac{1}{6} a + 3 - \frac{1}{3}a - 5 = - \frac{a}{6} - 2$

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3 years ago

Choose the correct simplification of the expression f to the 6th power times h to the 9th power all over f to the 3rd power time

maria [59]

That doesn't make sense

6 0

4 years ago

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37. One half of the water is poured out of a full container. Then one third of the remainder is poured out. Continue the process

ratelena [41]

After the first pour the water remains 1/2, after the second 1/3, after the third is 1/4, and after 9 pour the water remains 1/10.

<h3>What is a fraction?</h3>

Fraction number consists of two parts, one is the top of the fraction number which is called the numerator and the second is the bottom of the fraction number which is called the denominator.

After the first pour container has = 1/2

After the second pour = 1/3

Third pour = 1/4

After n pour = 1/(n+1)

n = 9

= 1/10

Thus, after the first pour the water remains 1/2, after the second 1/3, after the third is 1/4, and after 9 pour the water remains 1/10.

Learn more about the fraction here:

brainly.com/question/1301963

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5 0

2 years ago

H(x) = x2 + 1 k(x) = x – 2 (h + k)(2) = (h – k)(3) = Evaluate 3h(2) + 2k(3) =

matrenka [14]

For this case we have the following functions:

$h (x) = x ^ 2 + 1$

$k (x) = x - 2$

We must perform each of the operations shown.

We have then:

For (h + k) (2):

$(h + k) (x) = h (x) + k (x)$

$(h + k) (x) = (x ^ 2 + 1) + (x - 2)$

$(h + k) (x) = x ^ 2 + x - 1$

Evaluating for x = 2 we have:

$(h + k) (2) = 2 ^ 2 + 2 - 1$

$(h + k) (2) = 4 + 2 - 1$

$(h + k) (2) = 5$

For (h - k) (3):

$(h - k) (x) = h (x) -k (x)$

$(h - k) (x) = (x ^ 2 + 1) - (x - 2)$

$(h - k) (x) = x ^ 2 - x + 3$

Evaluating for x = 3 we have:

$(h - k) (3) = 3 ^ 2 - 3 + 3$

$(h - k) (3) = 9 - 3 + 3$

$(h - k) (3) = 9$

For 3h (2) + 2k (3):

$3h (2) + 2k (3) = 3 (2 ^ 2 + 1) + 2 (3 - 2)$

$3h (2) + 2k (3) = 3 (5) + 2 (1)$

$3h (2) + 2k (3) = 15 + 2$

$3h (2) + 2k (3) = 17$

Answer:

$(h + k) (2) = 5$

$(h - k) (3) = 9$

$3h (2) + 2k (3) = 17$

4 0

3 years ago

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