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shepuryov [24]
3 years ago
13

The formula for the area of a triangle is A= bh, where b is the base of the triangle and h is the height of the triangle. What i

s the length of
the base if the area is 32 cm and the height is 4 cm?
A 4 cm
B. 8 cm
C. 16 cm
D. 18 cm
Mathematics
1 answer:
vazorg [7]3 years ago
8 0
8cm cuase 32 is how área and your look for the base(b) and height (h) so you multiply 4 witch is your (h) and your lookin for you (b) and is going to be 8 cuase 4 times 8 is 32 witch is you area (a) so the answer is 8 cm for you (b)
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Find the greatest common factor of the following monomials 26n5 12n2
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The greatest common factor is 2n^{2}


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With the variables, we take out as many as the lowest number will let us. Since the smallest number of n's is 2 in the second term, we take that many.

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Cuantos diez hay en el numero 568?​
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Hay 56 diez en el numero 568.

Step-by-step explanation:

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4 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}

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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
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Answer:

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sin tita + sin ^2 tita = 0

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