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koban [17]
3 years ago
10

What is the fourth term in the binomial expansion of (2x – y) 7?

Mathematics
1 answer:
nata0808 [166]3 years ago
8 0

Answer:

(7C4) (2x)^4 (-y)^{7-4}

And replacing we got:

35 (2^4) x^4 (-y)^{-3}

And then the final term would be:

-560 x^4 y^3

Step-by-step explanation:

For this case we have the following expression:

(2x-y)^7

And we can use the binomial theorem given by:

(x+y)^n =\sum_{k=0}^n (nCk) x^k y^{n-k}

And for this case we want to find the fourth term and using the formula we have:

(7C4) (2x)^4 (-y)^{7-4}

And replacing we got:

35 (2^4) x^4 (-y)^{-3}

And then the final term would be:

-560 x^4 y^3

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A window is twice as long as it is wide. The area of the window is 2x where x
Alchen [17]

Answer:

When we have a rectangle of length L and width W, the area is calculated as:

A = W*L

In this case, W = x, and the area is:

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We know that the area of the window and the frame (together) is:

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The area of the frame alone, will be equal to the difference between the area of both objects (a) and the area of the window alone (A)

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5 0
3 years ago
Solve this A tourist starts off from town A and travels for 50km on a bearing of N80°W to town B.At town B ,he continues for ano
VLD [36.1K]

Answer:

The point C is 12.68 km away from the point A on a bearing of S23.23°W.

Step-by-step explanation:

Given that AB is 50 km and BC is 40 km as shown in the figure.

From the figure, the length of x-component of AC = |AB sin 80° - BC cos 20°|

=|50 sin 80° - 40 cos 20°|=11.65 km

The length of y-component of AC = |AB cos 80° - BC sin 20°|

=|50 cos 80° - 40 sin 20°|= 5 km

tan\theta= 5/11.65

\theta=23.23°

AC= \sqrt{5^2+11.65^2}=12.68 km

Hence, the point C is 12.68 km away from the point A on a bearing of S23.23°W.

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elena55 [62]
It might be D in my opinion? But I probably did my work wrong.. sorry!
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