How many radians is 270°??

1 answer:

3 0

Converting degrees to radians is easy if you remember,
$\pi =180$
So anytime you have an angle, just divide by 180 and put pi on the end.
$270 =\ \textgreater \ \frac{270}{180} \pi = \frac{3}{2} \pi$

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Between which two consecutive integers does <img src="https://tex.z-dn.net/?f=%5Csqrt%7B%7D%2012%5C%5C" id="TexFormula1" title="

Annette [7]

The most appropriate choice for square root will be given by-

$\sqrt{12}$ lies between 3 and 4

What is square root of a number?

Number raised to the power $\frac{1}{2}$ is the square root of the number.

Here

$\sqrt{12}\\ 2\sqrt{3}\\ 2*1.732\\3.464\\$

3.464 lies between 3 and 4

$\sqrt{12}$ lies between 3 and 4

To learn more about square root of a number, refer to the link:

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1 year ago

How much do you need to add to 2 5/8 to make 6

docker41 [41]

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3 3/8

Step-by-step explanation:

6 minus 2 5/8 equals 3 3/8

3 3/8 plus 2 5/8 equals 6

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3 years ago

Read 2 more answers

Let O be an angle in quadrant III such that cos 0 = -2/5 Find the exact values of csco and tan 0.

vivado [14]

well, we know that θ is in the III Quadrant, where the sine is negative and the cosine is negative as well, or if you wish, where "x" as well as "y" are both negative, now, the hypotenuse or radius of the circle is just a distance amount, so is never negative, so in the equation of cos(θ) = - (2/5), the negative must be the adjacent side, thus

$cos(\theta)=\cfrac{\stackrel{adjacent}{-2}}{\underset{hypotenuse}{5}}\qquad \textit{let's find the \underline{opposite side}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{5^2 - (-2)^2}=b\implies \pm\sqrt{25-4}\implies \pm\sqrt{21}=b\implies \stackrel{III~Quadrant}{-\sqrt{21}=b}$

$\dotfill\\\\ csc(\theta)\implies \cfrac{\stackrel{hypotenuse}{5}}{\underset{opposite}{-\sqrt{21}}}\implies \stackrel{\textit{rationalizing the denominator}}{-\cfrac{5}{\sqrt{21}}\cdot \cfrac{\sqrt{21}}{\sqrt{21}}\implies -\cfrac{5\sqrt{21}}{21}} \\\\\\ tan(\theta)=\cfrac{\stackrel{opposite}{-\sqrt{21}}}{\underset{adjacent}{-2}}\implies tan(\theta)=\cfrac{\sqrt{21}}{2}$