Math, asked by himanshu1234579, 1 year ago

can anybody solve all this questions

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Answers

Answered by CoruscatingGarçon
0

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

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