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OverLord2011 [107]
3 years ago
5

BE bisects <ABC. m<ABE=44. Find m<CBE.

Mathematics
1 answer:
boyakko [2]3 years ago
6 0

Answer:

The measure of ∠CBE is 44° ⇒ a

Step-by-step explanation:

The bisector of any angle divides the angle into two congruent angles (two angles equal in measures)

∵ BE bisects ∠ABC

∴ BE divides ∠ABC into two angles, ∠ABE and ∠CBE

→ By using the rule above

∵ ∠ABE ≅ ∠CBE

∴ m∠ABE = m∠CBE

∵ m∠ABE = 44°

∵ m∠ABE = m∠CBE

∴ m∠CBE = 44°

∴ The measure of ∠CBE is 44°

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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:

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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

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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

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