Solve using factoring

Mathematics

1 answer:

jolli1 [7]3 years ago

7 0

The equation is balanced by the 3's on both sides so that cancels out. we are left with:
x^2-6x=0
we take out an x, leaving us with:
x(x-6)=0
we know that to find the solution [ or "zero's", ] we must set "x"=0, and "x-6"=0.
the solutions are: x=0 and 6.
hope that helped, xx

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Given that θ terminates in Quadrant III and tanθ 5/12= , find cscθ

Komok [63]

$\bf tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
csc(\theta)=\cfrac{hypotenuse}{opposite}
\\\\
-------------------------------\\\\
tan(\theta)=\cfrac{5}{12}\cfrac{\leftarrow opposite=b}{\leftarrow adjacent=a}\\\\
-------------------------------\\\\
\textit{let's use the pythagorean theorem to get the hypotenuse "c"}
\\\\\\
c^2=a^2+b^2\implies c=\pm\sqrt{a^2+b^2}\implies c=\pm\sqrt{12^2+5^2}
\\\\\\
\boxed{c=\pm 13}$

now, the square root gives us the +/- version, so.. which is it? well, the hypotenuse is just a radius unit, is never negative, just the radius unit, so just the absolute value of that or the version 13, so c = 13

now, that we know what the hypotenuse "c" is, well

$\bf csc(\theta)=\cfrac{hypotenuse}{opposite}\implies csc(\theta)\cfrac{13}{5}$