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ExtremeBDS [4]
3 years ago
15

Tracie ran a total of 5 3/4 miles on Saturday and Sunday. She ran 1 5/8 miles on Saturday . How many miles did Tracie run on Sun

day ?
Mathematics
1 answer:
mylen [45]3 years ago
8 0

Tracie ran \frac{33}{8} miles or 4\frac{1}{8} miles on sunday

<u><em>Solution:</em></u>

Given that Tracie ran a total of 5\frac{3}{4} miles on Saturday and Sunday

She ran 1\frac{5}{8} miles on Saturday

To find: Number of miles ran on Sunday

From given question,

Total miles ran on Saturday and Sunday = 5\frac{3}{4} miles

\rightarrow 5\frac{3}{4} = \frac{4 \times 5+3}{4} = \frac{23}{4} \text{ miles}

Miles ran on Saturday = 1\frac{5}{8} miles

\rightarrow 1\frac{5}{8} = \frac{8 \times 1+5}{8}=\frac{13}{8} \text{ miles }

Number of miles ran on sunday = Total miles ran on Saturday and Sunday - Miles ran on Saturday

Number of miles ran on sunday = \frac{23}{4} -\frac{13}{8}

\rightarrow \frac{23}{4} -\frac{13}{8}\\\\\rightarrow \frac{23 \times 2}{4 \times 2}-\frac{13}{8}\\\\\rightarrow \frac{46}{8} - \frac{13}{8}\\\\\rightarrow \frac{46-13}{8} = \frac{33}{8}

\rightarrow \frac{33}{8} = 4\frac{1}{8}

Thus Tracie ran \frac{33}{8} miles or 4\frac{1}{8} miles on sunday

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Area of Un-shaded Sector is 376.99

Step-by-step explanation:

From the information given, we can say:

Radius of Circle = 12 (Since KL is radius and KL = 12)

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Total circle is 360, so unshaded sector is:

360 - 60 = 300 degrees

So, to find Non-Shaded Sector Area, we use sector area formula:

A=\frac{\theta}{360}*\pi r^2

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\theta  is the angle of the sector (in our case, it is 300)

So we substitute and find the answer:

A=\frac{300}{360}*\pi (12)^2\\A=\frac{5}{6}*\pi*144\\A=120\pi\\A=376.99

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In a right triangle, we haev some trigonometric relationships between the sides and angles. Given an angle, the ratio between the opposite side to the angle by the hypotenuse is the sine of this angle, therefore, the following statement

\sin (\theta)=\frac{3}{5}

Describes the following triangle

To find the missing length x, we could use the Pythagorean Theorem. The sum of the squares of the legs is equal to the square of the hypotenuse. From this, we have the following equation

x^2+3^2=5^2

Solving for x, we have

\begin{gathered} x^2+3^2=5^2 \\ x^2+9=25 \\ x^2=25-9 \\ x^2=16 \\ x=\sqrt[]{16} \\ x=4 \end{gathered}

The missing length of the first triangle is equal to 4.

For the other triangle, instead of a sine we have a tangent relation. Given an angle in a right triangle, its tanget is equal to the ratio between the opposite side and adjacent side.The following expression

\tan (y)=\frac{12}{5}

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Using the Pythagorean Theorem again, we have

5^2+12^2=h^2

Solving for h, we have

\begin{gathered} 5^2+12^2=h^2 \\ 25+144=h^2 \\ 169=h^2 \\ h=\sqrt[]{169} \\ h=13 \end{gathered}

The missing side measure is equal to 13.

Now that we have all sides of both triangles, we can construct any trigonometric relation for those angles.

The sine is the ratio between the opposite side and the hypotenuse, and the cosine is the ratio between the adjacent side and the hypotenuse, therefore, we have the following relations for our angles

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\begin{gathered} \sin (\theta+y) \\ \cos (\theta+y) \end{gathered}

We can use the following identities

\begin{gathered} \sin (A+B)=\sin A\cos B+\cos A\sin B \\ \cos (A+B)=\cos A\cos B-\sin A\sin B \end{gathered}

Using those identities in our problem, we're going to have

\begin{gathered} \sin (\theta+y)=\sin \theta\cos y+\cos \theta\sin y=\frac{3}{5}\cdot\frac{5}{13}+\frac{4}{5}\cdot\frac{12}{13}=\frac{63}{65} \\ \cos (\theta+y)=\cos \theta\cos y-\sin \theta\sin y=\frac{4}{5}\cdot\frac{5}{13}-\frac{3}{5}\cdot\frac{12}{13}=-\frac{16}{65} \end{gathered}

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