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3 years ago

Find the slope and y intercept of the line y=7/5x-3 5/7; 3 3; 7/5 7/5;-3 -3; 7/5

Mathematics

2 answers:

7nadin3 [17]3 years ago

8 0

Step-by-step explanation:

Equation of a line is y = mx + c

where

m is the slope

c is the y intercept

From the question

y = 7/5x - 3

Comparing with the above formula

<h3>Slope / m = 7/5</h3><h3>c/ y intercept = - 3</h3>

Hope this helps you

Evgesh-ka [11]3 years ago

5 0

Answer:

slope is 7/5

y-intercept is -3

Step-by-step explanation:

y=mx+b

where

m = slope

b = y-intercept

so for

y = 7/5x-3

slope is 7/5

y-intercept is -3

(third option)

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Solve equation 3x-5y=7 5x-2y=-1 by elimination

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3 years ago

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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

A triangle with vertices at A(0, 0), B(0, 4), and C(6, 0) is dilated to yield a triangle with vertices at A′(0, 0), B′(0, 10), a

svlad2 [7]

A triangle ABC is similar to a triangle A´B´C´ The scale factor of the dilatation : s.
0 * s =0
4 * s = 10
6 * s = 15
s= 10/4=15/6=2.5
The scale factor of the dilatation is 2.5. It is shown in the attachment.

Download docx

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3 years ago