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masha68 [24]
4 years ago
7

How do you write 2x+2y=8 in slope-intercept form?

Mathematics
1 answer:
GuDViN [60]4 years ago
6 0
Subtract 2x from both sides of the equation (2y=-2x+8) , divide the entire equation by 2 to find y. These makes the equation y=-x+4
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This is a 10 point question i need help
Bess [88]
It’s B because they both equal each other I would explain more but I’m in class but the answer is definitely B
4 0
3 years ago
What is the equation of this line in standard form
Anestetic [448]

Answer:

6x-11y=-13

Step-by-step explanation:

(x1,y1)=(3/2,2) and (x2,y2)=(-4,-1)

y-y1= y2-y1/x2-x1 (x-x1)

y-2=-3/-11/2(x-3/2)

=6/11(x-3/2)

11(y-2)=6x-9

11y-22=6x-9

6x-11y=9-22

6x-11y=-13

7 0
3 years ago
Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

y^2=4-x^2-z^2

x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz

Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

I gotta go so I'll stop here, sorry.

7 0
3 years ago
Read 2 more answers
There is no school 4 the rest of the week 4 me STOP ASKING ALL THEM QUESTIONS
Leviafan [203]

Answer: lol why would you ask a question if it is not about school homework

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
What is the value of x?
Contact [7]
Using the Pythagoras theorem 

(sqrt 117)^2 = x^2 + 6^2
x^2 = sqrt117)^2 + 6^2
x^2 = 117 - 36 = 81

so x = 9 answer 
8 0
3 years ago
Read 2 more answers
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