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Vladimir79 [104]

3 years ago

15

Whats the domain of this function?

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:

6 0

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

6 0

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>

m∠RQU = 96 - 11

m∠RQU = 85°

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

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Evgesh-ka [11]

Answer:

72

Step-by-step explanation:

6 0

2 years ago

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