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Molodets [167]
3 years ago
15

A simplified form of 4x-4y 3x

Mathematics
1 answer:
Pavel [41]3 years ago
4 0
7x-4y just combine all the x and all the y sice both the x here are positive you add them and there is only one y so just keep it as it is
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1. What is the equation of a line that has a slope of ⅔ and a y-intercept of -3?
ArbitrLikvidat [17]

Answer:

1. y = (⅔)x - 3

2. y = 3x + c

3. 1) non-proportional

2) can be proportional if c = 0

Step-by-step explanation:

1. What is the equation of a line that has a slope of ⅔ and a y-intercept of -3?

y = (⅔)x - 3

2. What is the equation of a line that has a slope of 3?

y = 3x + c

3. Lable the 2 equations as proportional or non-proportional and why.

A proportional relation should pass through the origin, i.e the y-intercept should be 0

3 0
3 years ago
What the answer? 8x-(2x-3)=12
nevsk [136]

Answer:

x= 3/2 or 1.5

Step-by-step explanation:

First of all, you can take out the parenthesis because 8x is subtracting 2x-3.

8x-2x-3=12

6x-3=12

  +3  +3

6x= 15

6x/6= 15/6

x= 3/2 or 1.5

Hope this helps!

8 0
2 years ago
What is the volume of a sphere that has a radius of 6
Leya [2.2K]

Answer:288pi

Step-by-step explanation:4/3pir^3 6^3*4/3*pi 216*4/3*pi 288pi

8 0
3 years ago
Read 2 more answers
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
2 years ago
Read 2 more answers
Find the unknown number 5/12+()=1/2
xenn [34]

Hello from MrBillDoesMath!

Answer:    1/12


Discussion:

We solve  5/12 + x = 1/2 for x. Begin by subtracting 5/12 from each side

5/12 - 5/12 + x  =    1/2 - 5/12

0                  +x =    6/12 - 5/12      (as 1/2 = 6/12)

So x = 1/12

Regards, MrB

8 0
3 years ago
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