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astraxan [27]
3 years ago
11

What is the third term of (2x-1)^5

Mathematics
2 answers:
Ne4ueva [31]3 years ago
8 0
<span>1* (2x)^5 + 5*(2x)^4 * (-3)^1 + 10*(2x)^3 * (-3)^2 + 10*(2x)^2 * (-3)^3 + 5*(2x)^1 * (-3)^4 + 1 *(-3)^5
the third term would then be 720*x^3
I hope i helped you :D
</span>
Sergio039 [100]3 years ago
6 0
I hope this helps you

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Answer:

AE=22.4

Step-by-step explanation:

BE is 1/2 of BC

BC is 20 cm                 All sides of a square are equal

BE = 1/2 BC                  Property of a midpoint.

BE = 10

Now just use Pythagorus

AB^2 + BE^2 = AE^2

AE^2 = 20^2 + 10^2   Perform the sqrs

AE^2 = 400 + 100      Add the terms

AE^2 = 500                Take the square root of both sides

√AE^2 = √500

AE = 22.36

AE ≈ 22.4

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