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Vladimir79 [104]
3 years ago
15

Whats the domain of this function?

Mathematics
1 answer:
nika2105 [10]3 years ago
6 0

Answer:

to reflect the line and record readings

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Beth works as a sales rep and receives $43 for every $210 she sells. What is her rate of commission? Round to the nearest tenths
Soloha48 [4]

Answer: it should be 5 if im not mistaken

Step-by-step explanation:

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3 years ago
PLZZZ HELP ME OUT THIS DUE RIGHT NOW AND I FREAKING OUT PLEASE HELP ME P.S YOU HAVE WRITE ANSWERS FOR ALL OF THEM
Veseljchak [2.6K]
1. Negative
Explanation: A negative minus a positive will be a negative
2. Negative
Explanation: If you use your order of operations a negative multiplied by a negative will be positive but since BxB would be greater than A it would be negative for example 2-8 would equal -6
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3 years ago
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Pls help Im very stuck on this and don’t understand
Lady bird [3.3K]

Answer:

C. 85 degrees

Step-by-step explanation:

Step 1: Set up equation

x + 57 + 96 + x = 131

Step 2: Solve for <em>x</em>

2x + 153 = 131

2x = -22

x = -11

Step 3: Find m∠RQU

m∠RQU = 96 + <em>x</em>

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8 0
3 years ago
According to data from a medical​ association, the rate of change in the number of hospital outpatient​ visits, in​ millions, in
GaryK [48]

Answer:

a) f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient​ visits, in​ millions, is given by:

f'(t)=0.001155t(t-1980)^{0.5}

a) To find the function f(t) you integrate f(t):

\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt

To solve the integral you use:

\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}

Next, you replace in the integral:

\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C

Then, the function f(t) is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient​ visits.

Hence C' = 264,034,000

The function is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) For t = 2015 you have:

f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7

3 0
3 years ago
Help plssssssssssssssssss
Evgesh-ka [11]

Answer:

72

Step-by-step explanation:

6 0
2 years ago
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