Question

Evaluate the following pic

Answer 1

1)
`√(1225)+√(1024)=67`

2)
`\sqrt[3]{-1331}=-11`

3) Evaluating
`2:p :: p:8` we get
`p=\pm 4`

4)
`x^3+y^2+z \ when \ x=3, y=-2, x=-6 \ we \ get \ \mathbf{25}`

5)
`((-6)^4*(-2)^3*(3)^3)/((-6)^6)=-6`

Explanation:

1)
`√(1225)+√(1024)`

Prime factors of 1225 : 5x5x7x7

Prime factors of 1024: 2x2x2x2x2x2x2x2x2x2

`√(1225)+√(1024)\\=√(5*5*7*7)+√(2*2*2*2*2*2*2*2*2*2)\\=√(5^2*7^2)+√(2^2*2^2*2^2*2^2*2^2)\\=5*7+(2*2*2*2*2)\\=35+32\\=67`

`√(1225)+√(1024)=67`

2)
`\sqrt[3]{-1331}`

We know that
`\sqrt[n]{-x}=-\sqrt[n]{x} \ ( \ if \ n \ is \ odd)`

Applying radical rule:

`\sqrt[3]{-1331}\\=-\sqrt[3]{1331} \\=-\sqrt[3]{11*\11*11}\\=-\sqrt[3]{11^3} \\Using \ \sqrt[n]{x^n}=x \\=-11`

So,
`\sqrt[3]{-1331}=-11`

3)
`2:p :: p:8`

It can be written as:

`p*p=2*8\\p^2=16\\Taking \ square \ root \ on \ both \ sides\\√(p^2)=√(16)\\p=\pm 4`

Evaluating
`2:p :: p:8` we get
`p=\pm 4`

4)
`x^3+y^2+z \ when \ x=3, y=-2, x=-6`

Put value of x, y and z in equation and solve:

`x^3+y^2+z \\=(3)^3+(-2)^2+(-6)\\=27+4-6\\=25`

So,
`x^3+y^2+z \ when \ x=3, y=-2, x=-6 \ we \ get \ \mathbf{25}`

5)
`((-6)^4*(-2)^3*(3)^3)/((-6)^6)`

We know (-a)^n = (a)^n when n is even and (-a)^n = (-a)^n when n is odd

`((-6)^4*(-2)^3*(3)^3)/((-6)^6)\\\\=(1296*-8* 27)/(46656)\\\\=(-279936)/(46656) \\\\=-6`

So,
`((-6)^4*(-2)^3*(3)^3)/((-6)^6)=-6`