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sweet-ann [11.9K]
3 years ago
13

Factor -1.8 out of 3.6b-9

Mathematics
1 answer:
Fofino [41]3 years ago
7 0
<h3>Answer:   -1.8(-2b+5)</h3>

=================================================

Explanation:

Consider something like 2b+6 factoring to 2(b+3). When we distribute that outer 2 back inside the parenthesis, we're multiplying that 2 by everything inside. Factoring goes in reverse of this and we divide each term of 2b+6 by the GCF 2.

The same thing applies to this current problem as well.

Divide each term by the -1.8 we want to factor out.

  • (3.6b)/(-1.8) = -2b
  • (-9)/(-1.8) = 5

The results -2b and 5 will go inside the parenthesis. That's how we end up with -1.8(-2b+5)

You can use distribution to verify this

-1.8(-2b+5)

-1.8*(-2b) - 1.8*(5)

3.6b - 9

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6 0
2 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
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