Question

(80 points)
Please help! Show all of your work for your answers as well.

Answer 1

`\large\boxed{Q1.\ A=(5)/(6)x^2}\\\boxed{Q2.\ 250m^(12)n^(-3)=(250m^(12))/(n^3)}\\\boxed{Q3.\ (81x^(16)y^(36))/(16z^(28))}\\\boxed{Q5.\ y^(1)/(2)=√(y)}`

Explanation:

`Q1.\\\\\text{The formula of an area of a triangle:}\\\\A_\triangle=(bh)/(2)\\\\b-\ \text{base}\\h-\text{height}\\\\\text{We have}\ b=(5)/(3)x,\ h=x.\ \text{Substitute:}\\\\A_\triangle=(\left((5)/(3)x\right)(x))/(2)=(5x^2)/((3)(2))=(5)/(6)x^2`

`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)=(1)/(a^n)\\\\=(250m^(12))/(n^3)`

`Q3.\\\\\left(-(3x^4y^9)/(2z^7)\right)^4\qquad\text{use}\ \left((a)/(b)\right)^n=(a^n)/(b^n)\ \text{and}\ (ab)^n=a^nb^n\\\\=(3^4(x^4)^4(y^9)^4)/(2^4(z^7)^4)\qquad\text{use}\ (a^n)^m=a^(nm)\\\\=(81x^(4\cdot4)y^(9\cdot4))/(16z^(7\cdot4))=(81x^(16)y^(36))/(16z^(28))`

`Q4.\\\\\left(x^0y^{(1)/(3)\right)^(3)/(2)\cdot x^0\qquad\text{use}\ a^0=1\ \text{for any value of}\ a\ \text{except}\ 0\\\\=\left(1\cdot y^(1)/(3)\right)^(3)/(2)\cdot1\qquad\text{use}\ (a^n)^m=a^(nm)\\\\=y^{\left((1)/(3)\right)\left((3)/(2)\right)}\qquad\text{cancel 3}\\\\=y^(1)/(2)\qquad\text{use}\ \sqrt[n]{a}=a^(1)/(n)\\\\=√(y)`