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frez [133]
4 years ago
13

Olivia has 48 slices of ham and 36 slices of cheese to make sandwiches.

Mathematics
1 answer:
serious [3.7K]4 years ago
8 0

The greatest number of sandwiches that Olivia can make is 12.

3 slices of cheese will be on each sandwich

<u>Solution:</u>

Given that, Olivia has 48 slices of ham and 36 slices of cheese to make sandwiches.  

We have to find what is the greatest number of sandwiches that Olivia can make? And how many slices of cheese will be on each sandwich?

Let the number of sandwiches be "n"

Then, number of slices of ham per sandwich =\frac{48}{n}

And, number of slices of cheese per sandwich =\frac{36}{n}

As slices can’t be fractions, "n" should be a factor of 48 and 36. So now let us find H.C.F of 48 and 36 as we want greatest number of sandwiches that Olivia can make.

The factors of 48 are: 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48

The factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, and 36

The common factors are those which appear on both factors of 48 and 36 are 1, 2, 4, 6, and 12.

Out of these common factors, 12 is the greatest

So, H.C.F of 48 and 36 is 12, which means n = 12

Then, number of cheese slices per sandwich =\frac{36}{12}=3

Hence, 12 sandwiches can be made with 3 slices of cheese per sandwich.

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Answer:

Option C

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Step-by-step explanation:

Given in the question,

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Let suppose that cost price percentage = 100%

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<h3>cost% + markup%  = selling%</h3><h3>100% + 30% = 130%</h3>

So percent markup selling price = 30 / 130 x 100

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Hence, 30% markup on cost price = 23.0769% markup on selling price.

4 0
3 years ago
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}}

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3 years ago
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