What is Z^6y^3/z^6y^4

Mathematics

1 answer:

Maslowich3 years ago

4 0

Answer:

The correct answer is

Z⁶Y³/Z⁶Y⁴ = 1/Y

Step-by-step explanation:

<u>Points to remember</u>

1). xᵃ * xᵇ = xᵃ⁺ᵇ

2). xᵃ/xᵇ = xᵃ⁻ᵇ

3). x° = 1

4). x⁻ᵃ = 1/xᵃ

The given expression is Z⁶Y³/Z⁶Y⁴

<u>To find the simplification of expression</u>

Z⁶Y³/Z⁶Y⁴ = Z⁶⁻⁶ * Y³⁻⁴

= Z° * Y⁻¹ = 1 * 1/Y = 1/Y

Therefore the value of given expression Z⁶Y³/Z⁶Y⁴ is 1/Y

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Find the area of the triangle ABC with the coordinates of A(10, 15) B(15, 15) C(30, 9).

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Check the picture below. so, that'd be the triangle's sides hmmm so let's use Heron's Area formula for it.

$~\hfill \stackrel{\textit{\large distance between 2 points}}{d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2}}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ (\stackrel{x_1}{10}~,~\stackrel{y_1}{5})\qquad (\stackrel{x_2}{15}~,~\stackrel{y_2}{15}) ~\hfill a=\sqrt{[ 15- 10]^2 + [ 15- 5]^2} \\\\\\ ~\hfill \boxed{a=\sqrt{125}} \\\\\\ (\stackrel{x_1}{15}~,~\stackrel{y_1}{15})\qquad (\stackrel{x_2}{30}~,~\stackrel{y_2}{9}) ~\hfill b=\sqrt{[ 30- 15]^2 + [ 9- 15]^2} \\\\\\ ~\hfill \boxed{b=\sqrt{261}}$

$(\stackrel{x_1}{30}~,~\stackrel{y_1}{9})\qquad (\stackrel{x_2}{10}~,~\stackrel{y_2}{5}) ~\hfill c=\sqrt{[ 10- 30]^2 + [ 5- 9]^2} \\\\\\ ~\hfill \boxed{c=\sqrt{416}} \\\\[-0.35em] ~\dotfill$

$\qquad \textit{Heron's area formula} \\\\ A=\sqrt{s(s-a)(s-b)(s-c)}\qquad \begin{cases} s=\frac{a+b+c}{2}\\[-0.5em] \hrulefill\\ a=\sqrt{125}\\ b=\sqrt{261}\\ c=\sqrt{416}\\ s\approx 23.87 \end{cases} \\\\\\ A\approx\sqrt{23.87(23.87-\sqrt{125})(23.87-\sqrt{261})(23.87-\sqrt{416})}\implies \boxed{A\approx 90}$