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3 years ago

We call the part or the remaining number in a percent problem the_____________

Mathematics

1 answer:

NeTakaya3 years ago

7 0

<span>We call the part or the remaining number in a percent problem the amount.</span>

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Translate into an algebraic expression:

Ber [7]

The algebraic expression is: x% * 100

<h3>How to translate the expression?</h3>

The statement is given as:

How much x% of sugar syrup can you make if you have 100 grams of sugar?

The amount of sugar is given as:

100 grams

So, the algebraic expression is:

x% * 100

Read more about algebraic expression at:

brainly.com/question/19245500

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2 years ago

Identify the equivalent expression for each of the Expressions below x + 3 square root

aev [14]

Answer:

m8 its (3x)^1/2

Step-by-step explanation:

4 0

3 years ago

Ghchdhdjfhfjfjfjfjfjdjdjdjf

Artyom0805 [142]

Answer:

iugyvhjbnkijogyutfvbh

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Function to inverse function f(x) =2x/3-17

ZanzabumX [31]

Answer:

Inverse function is $f^{-1}(x)=\frac{3}{2}x+\frac{51}{2}$ .

Step-by-step explanation:

Given function is $f\left(x\right)=\frac{2}{3}x-17$ .

Now we need to find the inverse of the given function.

Step 1: replace f(x) with y

$y=\frac{2}{3}x-17$

Step 2: switch x and y

$x=\frac{2}{3}y-17$

Step 3: solve for y

$x=\frac{2}{3}y-17$

$x+17=\frac{2}{3}y$

$\frac{3}{2}(x+17)=y$

$\frac{3}{2}x+\frac{51}{2}=y$

Hence required inverse function is $f^{-1}(x)=\frac{3}{2}x+\frac{51}{2}$ .

3 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