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3 years ago

Please help me if you know the answer. Id like fr you to explain. Not recommended but I would like it.

Mathematics

1 answer:

Triss [41]3 years ago

5 0

I think it is 1/9 but I am not sure

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I'm confused on how to simplify variable expressions...

gulaghasi [49]

Basically, simplifying variables is the same as regular equations (like 5-3=2 and 5x-3x=2x)

So 7x-x is the same as 7x-1x which equals 6x

And -3p+6p is the same thing as -3+6 but with a p. With this in mind, the answer is 3p

Hope this helps

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3 years ago

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Jessica needs to memorize words on a vocabulary list for Russian class. She has memorized 21 of the words, which is one-third of

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3 years ago

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Which expression is equivalent to 3m+1-m? 2+m-1+m 1+m 3m-1 3m

Over [174]

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2+m-1+m

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3m+1-m

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3 years ago

What is the value of X in the equation 8X - 2y=48,when y=4

Travka [436]

Answer:

X=7

Step-by-step explanation:

8x-2y=48

8x-2(4)=48

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8(7)-8=48

56-8=48

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3 years ago

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Suppose theta is an angle in the standard position whose terminal side is in quadrant 4 and cot theta = -6/7. find the exact val

zimovet [89]

First off, let's notice that the angle is in the IV Quadrant, where sine is negative and the cosine is positive, likewise the opposite and adjacent angles respectively.

Also let's bear in mind that the hypotenuse is never negative, since it's simply just a radius unit.

$\bf cot(\theta )=\cfrac{\stackrel{adjacent}{6}}{\stackrel{opposite}{-7}}\qquad \impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{6^2+(-7)^2}\implies c=\sqrt{36+49}\implies c=\sqrt{85} \\\\[-0.35em] ~\dotfill$

$\bf tan(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{adjacent}{6}} ~\hfill csc(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{opposite}{-7}} ~\hfill sec(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{adjacent}{6}} \\\\\\ sin(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{sin(\theta)=\cfrac{-7}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies sin(\theta)=-\cfrac{7\sqrt{85}}{85}}$

$\bf cos(\theta)=\cfrac{\stackrel{adjacent}{6}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{cos(\theta)=\cfrac{6}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies cos(\theta)=\cfrac{6\sqrt{85}}{85}}$

6 0

3 years ago