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vagabundo [1.1K]
3 years ago
12

A pizzeria sold $200 worth of pepperoni pizza on Tuesday. On Wednesday, the pizzeria made 1/10 fewer sales than Tuesday how much

less money did the pizzeria make on pepperoni pizzas on Wednesday
Mathematics
2 answers:
Ivan3 years ago
4 0

Answer:

$20

Step-by-step explanation:

Since it's 1/10 fewer, you take 200 and take away a zero basically dividing it by ten making it 20

olya-2409 [2.1K]3 years ago
3 0

Answer: The pizzaria made $20 less on Wednesday than Tuesday.

1/10 of 200 is 20.

if you subtract 20 (the $20 less than Tuesday) from $200 you get $180, which is 9/10 of $200.

[this shows two different ways to find the answer, finding 1/10 or 9/10 of $200.]

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If the point (3,4) lies on the graph of the equation 3y=ax+7 ,find the value of a
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Find the height of a cuboid whose volume is 540cu.cm and base area is 90sq.cm
grandymaker [24]

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Step-by-step explanation:

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6 0
2 years ago
Help! Prove the equality<br><br>arccos √(2/3) - arccos (1+√6)/(2*√3) = π/6
stealth61 [152]

Answer:

<em>Proof in the explanation</em>

Step-by-step explanation:

<u>Trigonometric Equalities</u>

Those are expressions involving trigonometric functions which must be proven, generally working on only one side of the equality

For this particular equality, we'll use the following equation

\displaystyle cos(x-y)=cos\ x\ cos\ y+sin\ x\ sin\ y

The equality we want to prove is  

\displaystyle arccos\ \sqrt{\frac{2}{3}}-arccos\left(\frac{1+\sqrt{6}}{2\sqrt{3}}\right)=\frac{\pi}{6}  

Let's set the following variables:

\displaystyle x=arccos\ \sqrt{\frac{2}{3}},\ y=arccos(\frac{1+\sqrt{6}}{2\sqrt{3}})

And modify the first variable:

\displaystyle x=arccos\ \frac{\sqrt{6}}{3}}=>\ cos\ x= \frac{\sqrt{6}}{3}}

Now with the second variable

\displaystyle y=arccos\ \frac{1+\sqrt{6}}{2\sqrt{3}}=>cos\ y=\frac{1+\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{3}+3\sqrt{2}}{6}

Knowing that

sin^2x+cos^2x=1

We compute the other two trigonometric functions of X and Y

\displaystyle sin \ x=\sqrt{1-cos^2\ x}=\sqrt{1-(\frac{\sqrt{6}}{3})^2}=\sqrt{1-\frac{6}{9}}=\frac{\sqrt{3}}{3}

\displaystyle sin\ y=\sqrt{1-cos^2y}=\sqrt{1-\frac{(\sqrt{3}+3\sqrt{2})^2}{36}}}

\displaystyle sin\ y=\sqrt{\frac{36-(3+6\sqrt{6}+18)}{36}}=\sqrt{\frac{15-6\sqrt{6}}{36}}

Computing

15-6\sqrt{6}=(3-\sqrt{6})^2

Then

\displaystyle sin\ y=\frac{3-\sqrt{6}}{6}

Now we replace all in the first equality:

\displaystyle cos(x-y)=\frac{\sqrt{6}}{3}.\frac{\sqrt{3}+3\sqrt{2}}{6}+\frac{\sqrt{3}}{3}.\frac{3-\sqrt{6}}{6}

\displaystyle cos(x-y)=\frac{3\sqrt{2}+6\sqrt{3}}{18}+\frac{3\sqrt{3}-3\sqrt{2}}{18}

\displaystyle cos(x-y)=\frac{9\sqrt{3}}{18}=\frac{\sqrt{3}}{2}=cos\ \pi/6

Thus, proven  

5 0
3 years ago
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