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galina1969 [7]
3 years ago
5

5th term in expansion of (x-2)^7

Mathematics
1 answer:
jeka943 years ago
7 0

Answer:

560x^3

Step-by-step explanation:

The formula is

(r + 1)th term of (a + x)^n = nCr a^(n-r)x^r

So 5th term = (4 + 1)th term

= 7C4 x^(7-4) (-2)^4

=  35x^3*16

= 560x^3

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What is the prime factorizarion for 500​
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Answer:

2 exponents (2) times 5 exponents (3)

Step-by-step explanation:

To find the prime factors you start by dividing the number by the first prime number which is 2 if there is not a remainder meaning you can divide evenly anymore write down how many 2's you were able to divide by evenly now try dividing by the the next prime factor which is 3 the goal is to get to a quotient of 1

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How do I do 8b(ii) ? Please help me thank you!
Mila [183]
Step One
======
Find the length of FO (see below)

All of the triangles are equilateral triangles. Label the center as O
FO = FE = sqrt(5) + sqrt(2)

Step Two
======
Drop a perpendicular bisector from O to the midpoint of FE. Label the midpoint as J. Find OJ

Sure the Pythagorean Theorem. Remember that OJ is a perpendicular bisector.

FO^2 = FJ^2 + OJ^2
FO = sqrt(5) + sqrt(2)
FJ = 1/2 [(sqrt(5) + sqrt(2)]                                           \
OJ = ??

[Sqrt(5) + sqrt(2)]^2 = [1/2(sqrt(5) + sqrt(2) ] ^2 + OJ^2
5 + 2 + 2*sqrt(10) = [1/4 (5 + 2 + 2*sqrt(10) + OJ^2
7 + 2sqrt(10) = 1/4 (7 + 2sqrt(10)) + OJ^2      Multiply through by 4
28 + 8* sqrt(10) = 7 + 2sqrt(10) + 4 OJ^2    Subtract 7 + 2sqrt From both sides
21 + 6 sqrt(10) = 4OJ^2   Divide both sides by 4
21/4 + 6/4* sqrt(10) = OJ^2
21/4 + 3/2 * sqrt(10) = OJ^2 Take the square root of both sides.
sqrt OJ^2 = sqrt(21/4 + 3/2 sqrt(10) )
OJ = sqrt(21/4 + 3/2 sqrt(10) )

Step three
find h
h = 2 * OJ
h = 2* sqrt(21/4 + 3/2 sqrt(10) ) <<<<<< answer.


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