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3 years ago

What is the output value for the following function if the input value is 1?

Mathematics

1 answer:

Vladimir79 [104]3 years ago

7 0

Answer:

Output value for the following function is 4

Step-by-step explanation:

Given:

The input value are the value of x

$y = 3x + 1$

Input value is 1

We substitute the value 1 for the input variable x in the given function.

$y = 3x + 1$

$y = 3(1) + 1$

$y = 3 + 1$

$y = 4$

The output value are the value of y

Therefore, for an input of 1, we have an output of 4.

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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

List the 8 numbers between 4 and 40 in a sequence of 10 numbers a common difference

yuradex [85]

Answer:

8, 12, 16, 20, 24, 28, 32, 36

Step-by-step explanation:

the nth term of an arithmetic sequence is

$a_{n}$ = a₁ + (n - 1)d

a₁ is the first term and d the common difference

given a₁ = 4 and a₁₀= 40 , then

4 + 9d = 40 ( subtract 4 from both sides )

9d = 36 ( divide both sides by 9 )

d = 4

add d to each term to obtain the next term

4 + 4 = 8

8 + 4 = 12

12 + 4 = 16

16 + 4 = 20

20 + 4 = 24

24 + 4 = 28

28 + 4 = 32

32 + 4 = 36

the 8 terms between 4 and 40 are

8, 12, 16, 20, 24, 28, 32, 36

3 0

1 year ago