Question

o, Nhat is-the Sum of he Finite arthmche Seriks ST

Answer 1

Solution:

(10). The series in not an arithmetic series. There is some misprint.

(1). $\sqrt{625x^{20}y^8}$

$=\sqrt{25\times 25\times x^{10}\times x^{10}\times y^4\times y^4}$

$=\sqrt{(25)^2\times (x^{10})^2\times (y^4)^2}$

$=(25)\times (x^{10})\times (y^4)$

$=25x^{10}y^4$

(2).  

$\frac{\sqrt[3]{7}}{\sqrt[3]{49}}$

$\frac{\sqrt[3]{7}}{\sqrt[3]{7^2}}$

Multiplying by $\sqrt[3]{7}$ on nominator as well as denominator.

$=\frac{\sqrt[3]{7}\times \sqrt[3]{7}}{\sqrt[3]{7^2}\times \sqrt[3]{7}}$

$=\frac{\sqrt[3]{7\times 7}}{\sqrt[3]{7^2\times 7}}$

$=\frac{\sqrt[3]{49}}{\sqrt[3]{7^3}}$

$=\frac{\sqrt[3]{49}}{7}$

This is the simplified form.

(3).

$16=(x-2)^{\frac{2}{3}}$

Powering 3/2 on both sides

$16^{\frac{3}{2}}=[(x-2)^{\frac{2}{3}}]^{\frac{3}{2}}$

$(4^2)^{\frac{3}{2}}=[(x-2)^{\frac{2}{3}}]^{\frac{3}{2}}$

using the property, $(x^{a})^{b}=x^{a\times b}$ . SO,

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

$4^3=x-2$

$64=x-2$

$x=64+2$

$x=66$

(4).  

$f(x)=x^2+7$

$g(x)=\frac{x-3}{x}$

$(fog)(x)=f(\frac{x-3}{x})$

$(fog)(x)=(\frac{x-3}{x})^2+7$

$(fog)(x)=\frac{(x-3)^2}{x^2}+7$

$(fog)(x)=\frac{x^2+9-6x}{x^2}+7$

$(fog)(-3)=\frac{(-3)^2+9-6(-3)}{(-3)^2}+7$

$(fog)(-3)=\frac{9+9+18}{9}+7$

$(fog)(-3)=\frac{36}{9}+7$

$(fog)(-3)=4+7$

$(fog)(-3)=11$