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3 years ago

Write an equation of the line passing through the point A(2, 0) that is parallel to the line y = 3x−5

Mathematics

2 answers:

DIA [1.3K]3 years ago

8 0

Answer:

y = 3x - 6

Step-by-step explanation:

Parallel lines have the same slope,

So m = 3

y - 0 = 3(x - 2)

y = 3x - 6

satela [25.4K]3 years ago

4 0

Answer:

y = 3x-6

Step-by-step explanation:

y = 3x−5

This is in slope intercept form

y = mx+b where m is the slope and b is the y intercept

m =3

Parallel lines have the same slope

We have the slope m=3 and a point (2,0)

y = mx+b

y = 3x+b

Substituting the point into the equation to solve for b

0 = 3(2)+b

0 = 6+b

b = -6

y = 3x-6

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3 years ago

Read 2 more answers

1.A bag has 4 green blocks, 3 red blocks, 2 blue blocks, and 1 yellow block. Find the probability of randomly selecting a block

maks197457 [2]

Answer: 1) 7/10 2) 15 3) 240 4) 1/4 5) 4/9 6) 1/20 7) No, not independent

Step-by-step explanation:

1. 4 green+3red+2blue+1yellow = 10 total

if 3/10 are red, then 7/10 are not red. So the second option, 7/10

2. 3 cartons in every 40

40*5 = 200

so do 3*5 = 15

so 15 or the third option

3. 40 out of 110

110*6 = 660

so do 40*6 = 240

so 240 or the fourth option

Flipping heads: 1/2

rolling even: well, there is 2, 4, and 6, or 3 even numbers so 3/6 or 1/2

do 1/2*1/2 = 1/4

so 1/4 or the third option

5. Numbers more than 2: 3, 4, 5, 6 (4 numbers)

4/6 or 2/3 chance the first time, 2/3 chance the second time

do 2/3*2/3 = 4/9

so 4/9 or the second option

6. 5 dishes and 4 desserts

1/5 chance of picking the dish

1/4 chance of picking the dessert

do 1/5*1/4 = 1/20

so 1/20 or the fourth option

7. Since you don’t replace the marble after picking it, it affects the outcome of the second draw. Therefore the events are dependent, or not independent. So the second option.

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3 years ago

<img src="https://tex.z-dn.net/?f=5%28sin%28t%29%20%20%5Cfrac%7Bdy%7D%7Bdx%7D%20%2Bycos%28t%29%29%29%3Dcos%28t%29%20%20%28sin%28

Nataliya [291]

Assuming you mean

$5\sin t\dfrac{\mathrm dy}{\mathrm dt}+5y\cos t=\cos t\sin^2t$

This ODE is linear in $y$ , and you can already contract the left hand side as the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[5y\sin t\right]=\cos t\sin^2t$

Integrating both sides with respect to $t$ yields

$5y\sin t=\displaystyle\int\cos t\sin^2t\,\mathrm dt$
$5y\sin t=\dfrac13\sin^3t+C$
$y=\dfrac1{15}\sin^2t+C\csc t$

Given that $y\left(\dfrac\pi2\right)=9$ , we have

$9=\dfrac1{15}\sin^2\dfrac\pi2+C\csc\dfrac\pi2$
$9=\dfrac1{15}+C$
$C=\dfrac{134}{15}$

so that the particular solution over the interval is

$y=\dfrac1{15}\sin^2t+\dfrac{134}{15}\csc t$