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3 years ago

16. In the figure below, lines A and B are parallel.

Mathematics

1 answer:

lesantik [10]3 years ago

3 0

Answer:

the angle measurement is 75

Step-by-step explanation:

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The tax rate is 3.9% What is the tax on 42$? Round to the nearest hundredth

mafiozo [28]

To find the total price, we use this equation:
42 + 0.039(42)
We can make it simpler:
1.039(42)
Multiply:
43.638
Because we're rounding
The total price is $43.64

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3 years ago

2) f(x) = x2 - 6x +11 Degree: Lead Coefficient: End behavior:

Bad White [126]

Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

Over which interval is the graph of the parent absolute value function f(x)=|x| decreasing?

pogonyaev

Answer:

(–∞, 0)

Step-by-step explanation:

The graph of the absolute value parent function f(x) = |x| is shown in the attached diagram.

We can clearly see that from -∞ to 0, the function is decreasing and from 0 to +∞, it is increasing. This is the basic, parent absolute value function.

THe questions asks, when is it decreasing, so clearly, it is from -∞ to 0.

This is the answer.

7 0

3 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