Giving brainliest whoever answers this question first, please help

Please help me answer this question

Oliga [24]

Answer:

B

Step-by-step explanation:

7 0

3 years ago

Terri is rewriting 7 to the 13th power in expanded form. How many factors should she use in her expression?

Pachacha [2.7K]

13

Because the exponent is really just how many factors you have.
For example 2 to the power 5 would have 5 factors in expanded form.

7 0

3 years ago

Read 2 more answers

I beg someone to answer ive posted this 3 times and im still not sure how to do it.....

Anni [7]

Answer:

D) 3 X 10⁷

Step-by-step explanation:

<u>Estimated days:</u>

Jan 2000 to June 2001 = 1.5 years = 365*1.5 days ≈ 548 days

1 day = 24 hours = 24*60 minutes = 24*60*60 seconds =

86400 seconds

548 days = 548 * 86400 seconds =

47347200 ≈ 4.7*10⁷

Best estimate is D) 3 X 10⁷

8 0

3 years ago

Which function has a removable discontinuity?

sweet-ann [11.9K]

"f(x)=(5x)/(x-x²)" is the one function among the following choices given in the question that <span>has a removable discontinuity. The correct option among all the options that are given in the question is the third option or option "C". It is the only function having a term cancelling out each other at the top and the bottom. </span>

5 0

2 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