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yarga [219]
3 years ago
6

In the diagram below, DE is parallel to XY. What is the value of x?

Mathematics
1 answer:
iren2701 [21]3 years ago
7 0

Answer:

A. 85

Step-by-step explanation:

Since DE and XY are parallel, this means that the given angle and angle x are alternate Angles. According to the law of alternate Angles, this means that they are equal. It also makes sense as the angle x is the same angle as the angle directly opposite angle 85. This is because they are on the same line across parallel lines. This also means they are equal because vertically opposite angles are always equal.

Therefore x =85

Hope this helped!

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Guys, Can you please help me with these questions
MakcuM [25]

Answer:

a) w^{13} x^{5} y^{6}

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Step-by-step explanation:

a) (w^{2} xy^{3} )^{2}(w^{3}x )^{3}

1. Distribute the second power (2) outside the first pair of parenthesis:

(w^{2(2)} x^{2} y^{3(2)} )

= w^{4} x^{2} y^{6} (w^{3}x )^{3}

2. Distribute the third power (3) outside the second pair of parenthesis:

(w^{3(3)} x^{3} )

= w^{4} x^{2} y^{6} w^{9} x^{3}

3. Combine like terms:

w^{13} x^{5} y^{6}

--------------------------------------------

b) \frac{2x^{2} y^{5} }{6xy^{11} }

1. Factor the number 6 (= 2 · 3):

\frac{2x^{2} y^{5} }{2(3)xy^{11} }

2. Cancel the common factor (2):

\frac{x^{2} y^{5} }{3xy^{11} }

3. Cancel out xy^{5} in the numerator an denominator:

\frac{x}{3y^{6} }

hope this helps!

4 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.

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3 years ago
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kramer
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x = -4/2....reduces to -2

or this way...
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15x - 9x + 10 = -2 ...subtract 10 from both sides
6x + 10 - 10 = -2 - 10...simplify
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