Question

Q-3) Divide the following: * =
(i) -36m²n² by - 3m²n
(ii) 48x³y4 by 6x²y²
Q-4) If x = 2, then find the values of x² + and x² + - x2
Q-5) Simplify: (+)²×(¹-3)* × (²) ²​

Answer 1

Answer:

sorry but I don't know ( sorry )

Answer 2

Answer:

i) $\mathrm{Simplify}\:\frac{-36m^2n^2}{-3m^2n}:\quad 12n$

Steps:

$\frac{-36m^2n^2}{-3m^2n}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{-b}=\frac{a}{b}$

$=\frac{36m^2n^2}{3m^2n}$

$\mathrm{Cancel\:the\:common\:factor:}\:m^2$

$=\frac{36n^2}{3n}$

$\mathrm{Factor\:the\:number:\:}\:36=3\cdot \:12$

$=\frac{3\cdot \:12n^2}{3n}$

$\mathrm{Cancel\:the\:common\:factor:}\:3$

$=\frac{12n^2}{n}$

$=12n$

ii)

$\mathrm{Simplify}\:\frac{48x^3y^4}{6x^2y^2}:\quad 8xy^2$

Steps:

$\frac{48x^3y^4}{6x^2y^2}$

$\mathrm{Factor\:the\:number:\:}\:48=6\cdot \:8$

$=\frac{6\cdot \:8x^3y^4}{6x^2y^2}$

$\mathrm{Cancel\:the\:common\:factor:}\:6$

$=\frac{8x^3y^4}{x^2y^2}$

$=\frac{8xy^4}{y^2}$

$=8xy^2$

Q.4)

$2^2+2^2+\left(-2^2\right)=4$

Steps:

$2^2+2^2+\left(-2^2\right)$

$\mathrm{Follow\:the\:PEMDAS\:order\:of\:operations}$

$=2^2+2^2-4$

$=4+2^2-4$

$=4+4-4$

$=4$

Q-5)

$\left(+\right)^2\left(1-3\right)\times \:2^2=-8^2$

Steps:

$\left(+\right)^2\left(1-3\right)\times \:2^2$

$\mathrm{Subtract\:the\:numbers:}\:1-3=-2$

$=2^2\left(-2\right)^2$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$=-2^2\times \:2^2$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^b\times \:a^c=a^{b+c}$

$2\times \:2^2=\:2^{1+2}$

$=-\left(+\right)^2\times \:2^{1+2}$

$\mathrm{Add\:the\:numbers:}\:1+2=3$

$=-2^3^2$

$2^3=8$

$=-8^2$