Answer:
[tex]\large\boxed{Q1.\ A=\dfrac{5}{6}x^2}\\\boxed{Q2.\ 250m^{12}n^{-3}=\dfrac{250m^{12}}{n^3}}\\\boxed{Q3.\ \dfrac{81x^{16}y^{36}}{16z^{28}}}\\\boxed{Q5.\ y^\frac{1}{2}=\sqrt{y}}[/tex]
Step-by-step explanation:
[tex]Q1.\\\\\text{The formula of an area of a triangle:}\\\\A_\triangle=\dfrac{bh}{2}\\\\b-\ \text{base}\\h-\text{height}\\\\\text{We have}\ b=\dfrac{5}{3}x,\ h=x.\ \text{Substitute:}\\\\A_\triangle=\dfrac{\left(\frac{5}{3}x\right)(x)}{2}=\dfrac{5x^2}{(3)(2)}=\dfrac{5}{6}x^2[/tex]
[tex]Q2.\\\\2(5m^4n^{-1})^3\qquad\text{use}\ (ab)^n=a^nb^n\\\\=2(5^3)(m^4)^3(n^{-1})^3\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=(2)(125)(m^{4\cdot3})(n^{-1\cdot3})\\\\=250m^{12}n^{-3}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}\\\\=\dfrac{250m^{12}}{n^3}[/tex]
[tex]Q3.\\\\\left(-\dfrac{3x^4y^9}{2z^7}\right)^4\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\ \text{and}\ (ab)^n=a^nb^n\\\\=\dfrac{3^4(x^4)^4(y^9)^4}{2^4(z^7)^4}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=\dfrac{81x^{4\cdot4}y^{9\cdot4}}{16z^{7\cdot4}}=\dfrac{81x^{16}y^{36}}{16z^{28}}[/tex]
[tex]Q4.\\\\\left(x^0y^{\frac{1}{3}\right)^\frac{3}{2}\cdot x^0\qquad\text{use}\ a^0=1\ \text{for any value of}\ a\ \text{except}\ 0\\\\=\left(1\cdot y^\frac{1}{3}\right)^\frac{3}{2}\cdot1\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=y^{\left(\frac{1}{3}\right)\left(\frac{3}{2}\right)}\qquad\text{cancel 3}\\\\=y^\frac{1}{2}\qquad\text{use}\ \sqrt[n]{a}=a^\frac{1}{n}\\\\=\sqrt{y}[/tex]
