Please tell me solution of question 7,8,9,10 and 13 please answer me tomorrow is my xam
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Please tell me solution of question 7,8,9,10 and 13 please answer me tomorrow is my xam
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7.
you can keep it in fractional form or you can also solve it further in decimal form
8.
9.
10.
13.
Answer:
13 Q) 64 / 729
10Q) - 3
9Q) 11x^3+ 2x^2 - 5x -6
8 Q) 5m^2-2mn +8
7Q) 47/8
Step-by-step explanation:
13Q) Given (p/q) = (2/3)^2 +(6/7)^0
Hint: a^0 =1
(p/q)^3 = ((2/3)^2)^3
= (2/3)^6 [∵ ((a) ^m)^n = (a)^mn]
= 64 /729
10Q) [{(2^8) ×(3)^4} / { 8×(-2)^5×27}] = ( -1) (2)^ (8-5-3) × (3)^(4-3)
= - 3
9Q) Adding like terms we get
x^3(3+4+4) +x^2(-5+6+2-1) +x(4 -9)+2-8
= 11x^3+ 2x^2 - 5x -6
8Q) 4m^2 + 3mn +8 -( -m)^2-5mn = 5m^2-2mn +8
7Q) [ (3/4)÷(2/5) ]+ [ (6/5)×(10 /3)] = (3/4×5/2)×4 = 15/8 +4 = (15+32)/8 =47/8