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SVETLANKA909090 [29]
2 years ago
9

Factor the GCF: 30a3b4 + 20a2b2 - 15ab3. Please help

Mathematics
1 answer:
Blizzard [7]2 years ago
8 0

Answer:

The GCF for the variable part is ab².

GCF<em>Variable</em>=ab²

Multiply the GCF of the numerical part 5 and the GCF of the variable part ab².

5ab²

Step-by-step explanation:

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Step-by-step explanation: If you go through all the problems, just replace "b" with "(4)" into a calculator. For A and D, they both equaled 180.

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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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