I need help there is a stalker

Mathematics

1 answer:

stepan [7]2 years ago

3 0

Answer:

you really think anyone cares?

Step-by-step explanation:

Step 1- s t o p b e i n g a t r u m p i e

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Since most stores will calculate the total price of your purchase why is it important to be able to calculate discounts

kkurt [141]

Answer:

Step-by-step explanation:

it is important because if you don't then you will be charged more than what you are supposed to pay

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3 years ago

In the simplest terms that is equivalent to 2 1/3

ycow [4]

Answer:

IRREGULAR TERM OF FRACTIONS

Step-by-step explanation:

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= 7/3 ANSWER

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1 year ago

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Is sin theta=5/6, what are the values of cos theta and tan theta?

mina [271]

let's bear in mind that sin(θ) in this case is positive, that happens only in the I and II Quadrants, where the cosine/adjacent are positive and negative respectively.

$\bf sin(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{hypotenuse}{6}}\qquad \impliedby \textit{let's find the \underline{adjacent side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{6^2-5^2}=a\implies \pm\sqrt{36-25}\implies \pm \sqrt{11}=a \\\\[-0.35em] ~\dotfill$

$\bf cos(\theta )=\cfrac{\stackrel{adjacent}{\pm\sqrt{11}}}{\stackrel{hypotenuse}{6}} \\\\\\ tan(\theta )=\cfrac{\stackrel{opposite}{5}}{\stackrel{adjacent}{\pm\sqrt{11}}}\implies \stackrel{\textit{and rationalizing the denominator}~\hfill }{tan(\theta )=\pm\cfrac{5}{\sqrt{11}}\cdot \cfrac{\sqrt{11}}{\sqrt{11}}\implies tan(\theta )=\pm\cfrac{5\sqrt{11}}{11}}$