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shepuryov [24]
3 years ago
6

Work out(3/4) -2 (1/-3) = (_/_)​

Mathematics
1 answer:
Alenkinab [10]3 years ago
5 0

Answer:

3 1/12 as in the simplest form. And 37/12 as in improper fraction.

Step-by-step explanation:

(3/4) - 2 (1/-3)

= 3/4 + 2 1/3

= 9 / 12 + 2 4 / 12

= 2 13 / 12

= 2 + 1 1/12

= 3 1/12

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3 years ago
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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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katrin [286]

Answer:

1/3 of 12 would be greater than 1/4 of 12.

Step-by-step explanation:

One third (1/3) of twelve results in 4. (Multiply 12/1 by 1/3 and simplify.)

One fourth (1/4) of twelve results in 3. (Multiply 12/1 by 1/4 and simplify.)

4 > 3.

Therefore, 1/3 of 12 is greater then 1/4 of 12.

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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3 years ago
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