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10 months ago

Please help with this question

Mathematics

1 answer:

Slav-nsk [51]10 months ago

7 0

Answer:

A is correct. √11 and √14 are both between 3 and 4.

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I know i have asked a lot of questions but i am in need of help

viva [34]

the answer is 31/12

I hope this helps (;

5 0

3 years ago

Read 2 more answers

For his long distance phone service, Keith pays a $3 monthly fee plus 11 cents per minute. Last month, Keith's long distance bil

Allushta [10]

The number of minutes for which Keith was billed = 97

Step-by-step explanation:

Step 1:

It is given that Keith pays a fixed fee of 3$ a month and 11 cents per minute for the number of minutes used.

Total cost = Fixed cost + Per minute cost

If x represents the number of minutes consumed in a month then we can compute the total cost using the below equation:

Total cost = 3$ + 0.11 * x

Step 2:

The total cost is 13.67$

Substituting in the equation we get

13.67 = 3 + 0.11 * x

x = 10.67 / 0.11 = 97 minutes

Step 3:

Answer:

The number of minutes for which Keith was billed = 97

3 0

3 years ago

One size of pizza served at Joe’s Pizza Parlor is 10in. in diameter. What is the area of this particular (circular) size pizza?

abruzzese [7]

Answer:

78.5 in.²

Explanation:

Formula For Area Of Circle: Area = πr²

π: 3.14

r (Half Of Diameter): 5

r²: 25

3.14 · 25 = 78.5

6 0

2 years ago

Remainder of 42325/47=

m_a_m_a [10]

The answer is 900 r25

7 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