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11Alexandr11 [23.1K]
3 years ago
14

What is y=2x-3 y=-3/4x-1/4 with work shown please help

Mathematics
1 answer:
makvit [3.9K]3 years ago
8 0
2X-3 = -3/4x -1/4
Add 3 on both sides: 2X = -3/4x + 2.75
Add -3/4x on both sides: 2.75X = 2.75
Divide 2.75 on both sides: x= 1

hope this helped
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I NEED HELP ASAP PLS !!!
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Answer:

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Step-by-step explanation:

1.) 130

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OKay, i need help asap pls. The ratio__ is equivalent to 12/36. The ratio__is not equivalent to 12/36.
EleoNora [17]

The ratio of 1:3 is equivalent to 12/36. The ratio 3 / 4 is not equivalent to 12/36.

Given that,
The ratio__ is equivalent to 12/36. The ratio__is not equivalent to 12/36.
Blank statement to be filled from the options.

<h3>What is the Ratio?</h3>

The ratio can be defined as the comparison of the fraction of one quantity towards others. e.g.- water in milk.

Here,
12 / 36 = 1 / 3
and
12 / 36 ≠ 3 / 4

Thus,  the ratio of 1:3 is equivalent to 12/36. The ratio 3 / 4 is not equivalent to 12/36.


Learn more about Ratio here:

brainly.com/question/13419413

#SPJ1

The question seems to be incomplete,
So we will fill a ratio that is equivalent to 12 / 36 and one ratio that is not equivalent to 12 / 36,
We have two ratios here 1:3 and 3:4


8 0
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

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Answer:

The answer is 36.

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89+78=167

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