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Olin [163]
3 years ago
13

At Fruits R Us, Mr. Orchard realizes that dried pineapple slices and dried apricots were two of his top dried fruit sellers. So

he decided to make a mixture of the two. The dried pineapple slices sell for $5 per pound and the dried apricots sell for $8 per pound. If the store owner wants to make 300 pounds of the fruit mixture to sell at $7 per pound, how much of each type of fruit should be used to make the mixture.
Mathematics
1 answer:
Rudik [331]3 years ago
4 0

Answer:

Step-by-step explanation:

From the information given, you can write the following equations:

x+y=300 (1)

5x+8y=300*7

5x+8y=2100(2)

First, you can isolate x in (1):

x=300-y (3)

Now, you can replace (3) in (2):

5(300-y)+8y=2100

1500-5y+8y=2100

3y=2100-1500

y=600/3

y=200

Then, you can replace the value of y in (3) to find x:

x=300-200

x=100

According to this, the answer is that he should use 100 pounds of dried pineapple and 200 pounds of dried apricots.

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2 tan 30°<br>II<br>1 + tan- 300​
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Question:

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}

Answer:

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}= sin(60^{\circ})

Step-by-step explanation:

Given

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}

Required

Simplify

In trigonometry:

tan(30^{\circ}) = \frac{1}{\sqrt{3}}

So, the expression becomes:

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}

Simplify the denominator

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}

Express the fraction as:

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}= \frac{2}{\sqrt 3} / \frac{4}{3}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{2}{\sqrt 3} * \frac{3}{4}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{1}{\sqrt 3} * \frac{3}{2}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{3}{2\sqrt 3}

Rationalize

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{3\sqrt{3}}{2* 3}

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})} = \frac{\sqrt{3}}{2}

In trigonometry:

sin(60^{\circ}) =  \frac{\sqrt{3}}{2}

Hence:

\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}= sin(60^{\circ})

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