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3 years ago

25 out of 27 as a percentage

2 answers:

8 0

25/27 = 0.925925925...

It's almost equal to 1 and it turns out that 1 is 100%.

So, we'd say that it's equal to:

92.59% (to 2 decimal places)

We've multiplied the value above (25/27) by 100.

sweet-ann [11.9K]3 years ago

7 0

If you Divide 25 by 27 you get .9259

The first two numbers give you the percentage so its 92%

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9) 4 = K/42<br> Solve that for me pls I’m writing here because I need twenty words

DaniilM [7]

Answer:

k = 168

Step-by-step explanation:

$4= \frac{k}{42}$ rewrite the equation

$42 * 4=\frac{k}{42} *42$ Multiply 42 to isolate k. Do the same on the other side

42 * 4 = 168 When you do that you get 168.

8 0

3 years ago

Simplify: |3−8|−(12÷3+1)2

Nonamiya [84]

Answer:

|3-8|-(12/3+1)2

=|-5|-(4+1)2

=5-5x2

=5-10

=-5

You said simplify, but it can be solve.

7 0

3 years ago

Read 2 more answers

A shirt originally cost $36.24, but it is on sale for $30.80. What is the percentage decrease of the price of the shirt? If nece

KiRa [710]

To work out the decrease, do original amount - new amount and then take that answer and divide it by the original amount, and then multiply by 100.

So you would do:

36.24 - 30.80 = 5.44
5.44 ÷ 36.24 × 100 = 15% to the nearest percent.

6 0

3 years ago

4/12×1/8 in simplest form

Sliva [168]

Answer:

$\frac{1}{24}$

Step-by-step explanation:

$\frac{4}{12} *\frac{1}{8} =\frac{1}{24}$

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

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

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

4 0

2 years ago