All categories

2 years ago

Put -3x+2y=10 into y=mx+b

Mathematics

2 answers:

Goryan [66]2 years ago

4 0

Answer:

y = 3/2x + 5

Step-by-step explanation:

-3x + 2y = 10

+3x +3x Add 3x to both sides

2y = 3x + 10 Divide both sides by 2

y = 3/2x + 5

Natasha2012 [34]2 years ago

3 0

Answer:

y=3/2x+5

Step-by-step explanation:

You might be interested in

What does 170/100 equal in decimals?

astraxan [27]

It equals 1.7 because you're moving the decimal point by two places, when dividing by 100

4 0

3 years ago

The line L1 has equation 2x+y=8.The line L2 passes through the point A(7,4) and is perpendicular to L1.

baherus [9]

If m1 and m2 are the slopes of each line, for them to be perpendicular you gotta solve
m1 * m2 = -1

4 0

2 years ago

Is 34515 divisible my 9

ratelena [41]

Answer:

Yes, the answer would be 3835

Step-by-step explanation:

34515 divided by 9 is simply 3835

3 0

3 years ago

ASAP WHO EVER ANSWER THESE WILL BE MARKED BRANLIEST

valentinak56 [21]

Answer:

i dont know honestly

Step-by-step explanation:

3 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:

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

3 0

2 years ago