Question

Evaluate the following pic

Answer 1

Answer:

1) $\sqrt{1225}+\sqrt{1024}=67$

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) $\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}=-6$

Step-by-step explanation:

1) $\sqrt{1225}+\sqrt{1024}$

Prime factors of 1225 : 5x5x7x7

Prime factors of 1024: 2x2x2x2x2x2x2x2x2x2

$\sqrt{1225}+\sqrt{1024}\\=\sqrt{5\times5\times7\times7}+\sqrt{2\times2\times2\times2\times2\times2\times2\times2\times2\times2}\\=\sqrt{5^2\times7^2}+\sqrt{2^2\times2^2\times2^2\times2^2\times2^2}\\=5\times7+(2\times2\times2\times2\times2)\\=35+32\\=67$

$\sqrt{1225}+\sqrt{1024}=67$

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

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

Applying radical rule:

$\sqrt[3]{-1331}\\=-\sqrt[3]{1331} \\=-\sqrt[3]{11\times\11\times11}\\=-\sqrt[3]{11^3} \\Using \ \sqrt[n]{x^n}=x \\=-11$

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

3) $2:p :: p:8$

It can be written as:

$p*p=2*8\\p^2=16\\Taking \ square \ root \ on \ both \ sides\\\sqrt{p^2}=\sqrt{16}\\p=\pm 4$

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) $\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}$

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

$\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}\\\\=\frac{1296\times-8\times 27}{46656}\\\\=\frac{-279936}{46656} \\\\=-6$

So, $\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}=-6$