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Yanka [14]
3 years ago
9

Choose the simplified form of  the fifth term of 6C4(2x)2(-y2)4.​

Mathematics
1 answer:
Katarina [22]3 years ago
4 0

You didn't give us the choices.   It doesn't matter.  I think you're trying to write

\displaystyle {6 \choose 4} (2x)^2 (-y^2)^4

= \dfrac{6!}{4! 2!} ( 4x^2 y^8)

= \dfrac{6(5)}{2} ( 4x^2 y^8)

=60x^2 y^8

Answer: 60 x² y⁸

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7 0
2 years ago
The triangle T has vertices at (-2, 1), (2, 1) and (0,-1). (It might be an idea to
Firdavs [7]

Rewrite the boundary lines <em>y</em> = -1 - <em>x</em> and <em>y</em> = <em>x</em> - 1 as functions of <em>y </em>:

<em>y</em> = -1 - <em>x</em>  ==>  <em>x</em> = -1 - <em>y</em>

<em>y</em> = <em>x</em> - 1  ==>  <em>x</em> = 1 + <em>y</em>

So if we let <em>x</em> range between these two lines, we need to let <em>y</em> vary between the point where these lines intersect, and the line <em>y</em> = 1.

This means the area is given by the integral,

\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy

The integral with respect to <em>x</em> is trivial:

\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy

For the remaining integral, integrate term-by-term to get

\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line <em>y</em> = 1 and (0, -1)), so its area is 1/2*4*2 = 4.

6 0
3 years ago
What should the graph of two lines look like when there are no solutions that they have in common?
nataly862011 [7]
If the graphs of the equations do not intersect
4 0
2 years ago
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Ivahew [28]

Answer:

Step-by-step explanation:

The opposite side (the one not connected to A) = 4

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The adjacent side needs to be found for the cosine and the tangent.

a^2 + b^2 = c^2

a = opposite side = 4

b = adjacent side = ?

c = hypotenuse = 5

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16 + x^2 = 25          

x^2 = 25 - 16

x^2 = 9

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Tan(A) = opposite / adjacent = 4/3

cos(A) + tan(A) = 3/5 + 4/3

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4 0
3 years ago
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Make "a" the subject of:
slamgirl [31]

Answer:

a=m-x^2/k^2

Step-by-step explanation:

(k^2(m-a)) /x=x

K^2(m-a)=x^2

m-a=x^2/k^2

a=m-x^2/k^2

7 0
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