What is the fifth term of (2x + y)^8?

Expansion for there will be 9 terms.

(a+b) ^n =nco a^n b^o+ nc1 a^(n-1) b^1+nc2 a^(n-2) b^2+nc3 a(n-3) b^3+nc4  a^(n-4) b^4+ncn a^o b^n.

5th term of  (2x+y)^8 = ?  

5th term is nC4 a^(n-4) b^4

a=2x, b=y, n=8

therefore 8C4 (2x)^(8-4)  (y)^4

8!/((8-4)! 4!) (2x)^4 y^4

40320/576  (16 x^4 ) (y^4)

70 (16x^4)(y^4)

1120 X^4 Y^4

Step-by-step explanation:

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

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is there a picture

step-by-step explanation: