I really need help with Geometry

Mathematics

1 answer:

Rainbow [258]3 years ago

4 0

Can yu post the problem? my computer won't open the document

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X^5y^3/x^2-x-6 * x^2 -9/x^2y^4

zvonat [6]

$\bf \cfrac{x^5y^3}{x^2-x-6}\cdot \cfrac{x^2-9}{x^2y^4}\implies \cfrac{x^5y^3}{x^2-x-6}\cdot \cfrac{\stackrel{\textit{difference of squares}}{x^2-3^2}}{x^2y^4} \\\\\\ \cfrac{x^5y^3}{~~\begin{matrix} (x-3) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~(x+2)}\cdot \cfrac{~~\begin{matrix} (x-3) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~(x+3)}{x^2y^4}\implies \cfrac{x^5y^3}{x+2}\cdot \cfrac{x+3}{x^2y^4}$

$\bf \cfrac{x^5y^3}{x^2y^4}\implies \cfrac{x^5x^{-2}}{y^4y^{-3}}\cdot \cfrac{x+3}{x+2}\implies \cfrac{x^{5-2}}{y^{4-3}}\cdot \cfrac{x+3}{x+2}\implies \cfrac{x^3(x+3)}{y(x+2)}\implies \cfrac{x^4+3x^3}{yx+2y}$

6 0

4 years ago

use the change of base formula to evaluate log[3](58). then convert log[3](58) to a logarithm in base 4?

devlian [24]

Log[3](58)=3.69597450
log[3](58) to a logarithm in base 4= 167.957<span>
if you still need it.</span>

3 0

3 years ago

Any answer any one fast

scoray [572]

The answer should be A
as quadrilateral means polygons with 4 sides so the the triangle and the hexagon won't be
also none is parallelogram
and perpendicular lines means giving a angle of 90 degreess so no

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I really need help with Geometry​

I really need help with Geometry