Answer:
1) [tex]\sqrt{1225}+\sqrt{1024}=67[/tex]
2) [tex]\sqrt[3]{-1331}=-11[/tex]
3) Evaluating [tex]2:p :: p:8[/tex] we get [tex]p=\pm 4[/tex]
4) [tex]x^3+y^2+z \ when \ x=3, y=-2, x=-6 \ we \ get \ \mathbf{25}[/tex]
5) [tex]\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}=-6[/tex]
Step-by-step explanation:
1) [tex]\sqrt{1225}+\sqrt{1024}[/tex]
Prime factors of 1225 : 5x5x7x7
Prime factors of 1024: 2x2x2x2x2x2x2x2x2x2
[tex]\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[/tex]
[tex]\sqrt{1225}+\sqrt{1024}=67[/tex]
2) [tex]\sqrt[3]{-1331}[/tex]
We know that [tex]\sqrt[n]{-x}=-\sqrt[n]{x} \ ( \ if \ n \ is \ odd)[/tex]
Applying radical rule:
[tex]\sqrt[3]{-1331}\\=-\sqrt[3]{1331} \\=-\sqrt[3]{11\times\11\times11}\\=-\sqrt[3]{11^3} \\Using \ \sqrt[n]{x^n}=x \\=-11[/tex]
So, [tex]\sqrt[3]{-1331}=-11[/tex]
3) [tex]2:p :: p:8[/tex]
It can be written as:
[tex]p*p=2*8\\p^2=16\\Taking \ square \ root \ on \ both \ sides\\\sqrt{p^2}=\sqrt{16}\\p=\pm 4[/tex]
Evaluating [tex]2:p :: p:8[/tex] we get [tex]p=\pm 4[/tex]
4) [tex]x^3+y^2+z \ when \ x=3, y=-2, x=-6[/tex]
Put value of x, y and z in equation and solve:
[tex]x^3+y^2+z \\=(3)^3+(-2)^2+(-6)\\=27+4-6\\=25[/tex]
So, [tex]x^3+y^2+z \ when \ x=3, y=-2, x=-6 \ we \ get \ \mathbf{25}[/tex]
5) [tex]\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}[/tex]
We know (-a)^n = (a)^n when n is even and (-a)^n = (-a)^n when n is odd
[tex]\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}\\\\=\frac{1296\times-8\times 27}{46656}\\\\=\frac{-279936}{46656} \\\\=-6[/tex]
So, [tex]\frac{(-6)^4\times(-2)^3\times(3)^3}{(-6)^6}=-6[/tex]
