can anybody solve all this questions
Answers
Answer:
Q1
(3^-2 x 5^-1 x 4^4)/ -15^-3 x 6^-2
=(4^4 x -15^3 x 6^2)/(3^2 x 5^1)
={4^4 x -(5 x 3)^3 x (2 x 3)^2}/(3^2 x 5^1)
={4^4 x (-5)^3 x 3^3 x 2^2 x 3^2}/(3^2 x 5^1)
=2^(2*4+2) x (-5)^(3-1) x 3^(3-2)
=2^10 x (-5)^2 x 3^1
=1024 x 25 x 3
=76800
Q2
(3^-5 x 4^-3 x 7^-6 x a^-5)/(3^-3 x 4^-12 x 7^-4 x a^-2)
=(3^3 x 4^12 x 7^4 x a^2)/(3^5 x 4^3 x 7^6 x a^5)
=(3^(3-5) x 4^(12-3) x 7^(4-6) x a^(2-5))
=(3^-2 x 4^9 x 7^-2 x a^-3)
=4^9/(3^2 x 7^2 x a^3)
=262144 /441 x a^3
Q3
(5/9)^-7 / (5/9)^-2=(5/9)^2m-3
=(9/5)^7 / (9/5)^2=(9/5)^-(2m-3)
=(9/5)^7-2=(9/5)^-2m+3
=(9/5)^5=(9/5)^3-2m
As the bases are same, we can now consider the powers only,
=>5=3-2m
=>2m=3-5
=>2m=-2
=>m=-1
Q4
5^x-3 x 3^2x-8=225
5^x-3 x 3^2x-8=(3*5)^2
5^x-3 x 3^2x-8=3^2 x 5^2
As the bases are same, we can now consider the powers only(of 5),
x − 3 = 2
x=5
Q5
1)x^2+10x+21
= x^2 + 3x + 7x + 21
=x(x+3)+ 7(x+3)
=(x+7)(x+3)
2)x^2+7x+12
= x^2 + 4x + 3x + 12
= x (x + 4) + 3(x + 4)
= (x + 4)(x + 3)
3)5x^2+15x+10
=5 x ^2 +10 x+5x+10
=5x (x+2)+5(x+2)
= (x+2) (5x+5)
4)(x+y)^4-(x-y)^4
(x+y)^4 = x^4+4x^3y+6x^2y^2+4xy^3+y^4
(x-y)^4 = x^4-4x^3y+6x^2y^2-4xy^3+y^4
simplifying we get,
8x^3y + 8xy^3 = 8xy(x^2 + y^2)
8x^3y + 8xy^3-8xy(x^2 + y^2)
Q6
0.84-0.03x=2.78-x
-0.03x+x=2.78-0.84
0.97 x=1.94
x=2
Q7
(1(x-5)/3)=(1(x-3)/5)
By cross-multiplying
5(1(x-5))=3(1(x-3))
5x-25=3x-3
5x-3x=-3+25
2x=22
x=11
HOPE IT HELPS