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3 years ago

What is the equation of the line perpendicular to 3x+y=6 that passes through the point (-6, -1)?

Mathematics

1 answer:

lesya692 [45]3 years ago

5 0

Answer: y = -1/3x - 3

Step-by-step explanation:

First, convert this equation into slope intercept form (y = mx + b where m is the slope and b is the y-intercept)

To convert this equation into slope-intercept form, all you have to do is subtract 3x from both sides to give you the equation y = 3x + 6

Next, you need to find the slope of your new equation. The slope of a perpendicular line will have a slope that is the opposite reciprocal of the first slope, which means it will have the opposite sign and the fraction will be flipped (even whole numbers are a fraction - they are divided by 1)

Since the slope of the first equation is positive 3, you are going to change the sign to a negative and flip the fraction from 3/1 to 1/3, which gives you a slop of -1/3

Now that you have your slop, you need to find your y-intercept. To do this, you use the point that you are given. In the previous step, you figured out the slope, so now you can enter that into your new equation. Then, plug the x-coordinate in for x and the y-coordinate in for y in the equation.

At this point, your equation is y = -1/3x + b. You can plug -6 in for x and -1 in for y to give you the equation -1 = -1/3(-6) + b

The next step is to solve the equation like you would any other equation.

Multiply -6 by -1/3 to get 2, then subtract 2 from -1 to get b = -3

Lastly, create your new equation in slope-intercept form with the values you have found.

y = -1/3x -3

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2 years ago

Assume Y=1+X+u, where X, Y, and u=v+X are random variables, v is independent of X; E(v)=0, Var(v)=1, E(X)=1, and

kobusy [5.1K]

Answer:

a) $E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1+$

b) $E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2$

c) $E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2$

d) $E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6$

e) $E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1$

f) $E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3$

g) $E(u) = E(v) +E(X) = 0+1=1$

h) E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]

Step-by-step explanation:

For this case we know this:

$Y = 1+X +u$

$u = v+X$

with both Y and u random variables, we also know that:

$[tex] E(v) = 0, Var(v) =1, E(X) = 1, Var(X)=2$

And we want to calculate this:

Part a

$E(u|X=1)= E(v+X|X=1)$

Using properties for the conditional expected value we have this:

$E(u|X=1)= E(v|X=1) + E(X|X=1) = E(v) +1 = 0 +1 =1$

Because we assume that v and X are independent

Part b

$E(Y| X=1) = E(1+X+u|X=1)$

If we distribute the expected value we got:

$E(Y| X=1)= E(1|X=1) + E(X|X=1) + E(u|X=1) = E(1) + 1 + E(v) + 0 = 1+1+0=2$

Part c

$E(u|X=2)= E(v+X|X=2)$

Using properties for the conditional expected value we have this:

$E(u|X=2)= E(v|X=2) + E(X|X=2) = E(v) +2 = 0 +2 =2$

Because we assume that v and X are independent

Part d

$E(Y| X=2) = E(1+X+u|X=2)$

If we distribute the expected value we got:

$E(Y| X=2)= E(1|X=2) + E(X|X=2) + E(u|X=2) = E(2) + 2 + E(v) + 2 = 2+2+2=6$

Part e

$E(u|X) = E(v+X |X) = E(v|X) +E(X|X) = E(v) +E(X) = 0+1=1$

Part f

$E(Y|X) = E(1+X+u |X) = E(1|X) +E(X|X) + E(u|X) = 1+1+1=3$

Part g

$E(u) = E(v) +E(X) = 0+1=1$

Part h

E(Y) = E(1+X+u) = E(1) + E(X) +E(v+X) = 1+1 + E(v) +E(X) = 1+1+0+1 = 3[/tex]

8 0

3 years ago

PLEASE HELP! THANKS!

Vinvika [58]

So that is area of rectangle-1/2areacircle

so
areacircle=pir^2
r=5
1/2pir^2=1/2pi5^2=25pi/2=12.5pi

arearectangle=lw=12*5=60

unshaded=60-12.5pi=20.75 square feet
round

20.8 square feet

6 0

3 years ago