Round 0.42 to the nearst tenth

Mathematics

1 answer:

Archy [21]3 years ago

4 0

0.4 because if it’s five or more, add one more. If it’s four or less, let it rest.

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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.

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

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

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Answer:

The class width is 20

Step-by-step explanation:

In a frequency or a relative frequency distribution the class width is calculated as the difference between the lower or upper class limits of consecutive classes. A point to note is that all the categories or classes usually have the same class width.

We use the first two classes to calculate the class width by using their respective upper limits;

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Can someone please check my answer?!

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Answer:

You answered it correctly :)

Step-by-step explanation:

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