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3 years ago

5

HEELLLLPPPP!!!! If the inverses of two functions are both functions, will the inverse of the composite function made by the orig

inal functions also be a function?? I really need help

1 answer:

Anon25 [30]3 years ago

8 0

I pretty shure its true

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How do I do step by step

kiruha [24]

Answer:

1.1

Step-by-step explanation:

Write out a long division problem! Since you can't have a decimal in the divisor, multiply both numbers by 10. This will leave you with 39.6÷36. Now all you have to do is follow the steps and end up with an answer of 1.1

6 0

2 years ago

Findℒ{f(t)}by first using a trigonometric identity. (Write your answer as a function of s.)f(t) = 12 cost −π6

allsm [11]

Answer:

$L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ]$

Step-by-step explanation:

Given that:

$f(t) = 12 cos (t- \dfrac{\pi}{6})$

recall that:

cos (A-B) = cos AcosB + sin A sin B

∴

$f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}]$

$f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}]$

$f(t) = 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t)$

$L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ]$

$L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ]$

$L(f(t)) = 6 \sqrt{3} \dfrac{S}{S^2 + 1^2}+ 6 \dfrac{1}{S^2 +1^2}$

$L(f(t)) = \dfrac{6 \sqrt{3} +6 }{S^2+1}$

$L(f(t)) = \dfrac{6( \sqrt{3} \ S +1 }{S^2+1}$

$L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ]$

7 0

3 years ago

Evaluate gh- k when g= -6, h= 15, and k= -27.

Ivanshal [37]

Gh - k
<span>g= -6, h= 15, and k= -27
</span>
-6*15 - - 27
-90 + 27

-63 is the answer to the expression.

4 0

3 years ago

Read 2 more answers

What is the answer to 4(x + 1) = -6

snow_tiger [21]

Answer:

-5/2

Step-by-step explanation:

With Distributive property, we get 4x + 4 = -6. Subtract 4 from both sides and we get 4x = -10. Divide both sides by 4 and we get x = -5/2.

3 0

3 years ago

Read 2 more answers

A textbook company states that the average time a student needs to take a quiz from its book is 30 minutes with a standard devia

faltersainse [42]

Answer:

We conclude that her students, on average, do not take 30 minutes on the quiz, contrary to what the textbook company states.

Step-by-step explanation:

We are given;

Population mean; μ = 30 minutes

Population standard deviation; σ = 3 minutes

Sample size; n = 10

Sample mean; x¯ = 25

Significance level = 5% = 0.05

Let's define the hypotheses;

Null hypothesis; H0: μ = 30

Alternative hypothesis: Ha: μ ≠ 30

Let's find the test statistic;

z = (x¯ - μ)/(σ/√n)

z = (25 - 30)/(3/√10)

z = -5.27

From online p-value from z-score calculator attached using, z = -5.27, significance level = 0.05, two tailed hypothesis, we have;

p < 0.00001

This is less than the significance level, and so we will reject the null hypothesis and conclude that Her students, on average, do not take 30 minutes on the quiz, contrary to what the textbook company stated.

8 0

3 years ago

Read 2 more answers