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4 years ago

Write an equation of the line that is parallel to the line whose equation is 3y+7=2x and passes through point (2,6).

Mathematics

1 answer:

Sladkaya [172]4 years ago

5 0

Answer:

Step-by-step explanation:

To find the equation of a parallel line, we first need to find the slope of 3y + 7 = 2x.

We need to find it because, if two lines are parallel, then that means they have the same slope.

The best way to do this is solve for y:

$3y+7=2x\\3y=2x-7\\y=\frac{2}{3}-\frac{7}{3}$

Due to the rule that in y = mx + b, m = the slope, we find that the slope is 2/3.

Now we use that slope in the same formula to find b of the line we're trying to find the equation for, and then we'll have our answer. We find b by plugging in (2, 6) for x and y:

$y = mx+b\\y=\frac{2}{3}x+b\\6=\frac{2}{3}(2)+b\\18=4+3b\\ 14=3b\\\frac{14}{3} = b$

So our line is:

$y=\frac{2}{3}x+\frac{14}{3}$

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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:

$f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7$

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3 years ago