Wassup you know the square root for 320?

Mathematics

1 answer:

kupik [55]1 year ago

7 0

In order to calculate the square root of 320, first let's simplify it using the following property:

$\sqrt[c]{a^b}=a\sqrt[c]{a^{b-c}}$

So, factoring the number 320, we have:

$320\to2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot5=2^6\cdot5$

Now, calculating the square root and using

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How am I supposed to figure this out if csc Theta(0) has a radical number on the top?

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$\bf csc(\theta)=\cfrac{hypotenuse}{opposite}\implies csc(\theta)=\cfrac{\sqrt{7}}{2}\cfrac{\leftarrow hypotenuse=c}{\leftarrow opposite=b}\\\\
-----------------------------\\\\
\textit{so hmm using the pythagorean theorem, we can say that}\\\\
c^2=a^2+b^2\implies \pm \sqrt{c^2-b^2}=a\qquad 
\begin{cases}
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\textit{we'll assume is the + of the root, since you're not}\\
\textit{given further info on it, that means the I quadrant}\\\\
$

$\bf -----------------------------\\\\
cos(\theta)=\cfrac{a}{c}\qquad sec(\theta)=\cfrac{c}{a}\qquad tan(\theta)=\cfrac{b}{a}\qquad cot(\theta)\cfrac{a}{b}
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so just find the "adjacent" side, using the pythagorean theorem, and even though the square root can give you +/-, use the positive one, assuming the angle is in the 1st quadrant

keeping in mind, that the 2nd quadrant is feasible as well, since the opposite side of +2 can also be in the 2nd quadrant

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