What is 0.2 as a fraction

Mathematics

2 answers:

zaharov [31]3 years ago

5 0

Answer:

That is 1/5.

Step-by-step explanation:

It is also 20%(if you need that too.)

Vikentia [17]3 years ago

3 0

It’s 2/10 there you go since there’s no 0 infort of the 2 it’s always a 10

You might be interested in

Identify a pattern by making a table of the inputs and outputs.

-BARSIC- [3]

it wants you to make a table of the inputs and out puts, and what is the pattern that is happening. like is it a constant increase/decrease and things like that. i'll help you on the table...

x | y

1 | 5

2 | 4

3 | 3

4 | 2

(i am not sure what that 40 is at the end, but You get the point)

now all you have to do is identify the sequence or pattern you see here

3 0

3 years ago

Select all the statements that are true for any value of x

Goryan [66]

Answer:

A, C, E, F

Explanation:

equal on both sides

6 0

2 years ago

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

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

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