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3 years ago

Waylon played defense in 12 soccer games. This was 60% of the total games he played. How many games did Waylon play?

Mathematics

1 answer:

puteri [66]3 years ago

4 0

Answer:

20 games

Step-by-step explanation:

(12/x)=(60/100)

12*100=1200

60x=1200

x=20

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Write each expression as an algebraic (nontrigonometric) expression in u, u > 0.

max2010maxim [7]

Answer:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0$

Step-by-step explanation:

We want to write the trignometric expression:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)\text{ where } u>0$

As an algebraic equation.

First, we can focus on the inner expression. Let θ equal the expression:

$\displaystyle \theta=\sec^{-1}\left(\frac{u}{10}\right)$

Take the secant of both sides:

$\displaystyle \sec(\theta)=\frac{u}{10}$

Since secant is the ratio of the hypotenuse side to the adjacent side, this means that the opposite side is:

$\displaystyle o=\sqrt{u^2-10^2}=\sqrt{u^2-100}$

By substitutition:

$\displaystyle= \sin(2\theta)$

Using an double-angle identity:

$=2\sin(\theta)\cos(\theta)$

We know that the opposite side is √(u² -100), the adjacent side is 10, and the hypotenuse is u. Therefore:

$\displaystyle =2\left(\frac{\sqrt{u^2-100}}{u}\right)\left(\frac{10}{u}\right)$

Simplify. Therefore:

$\displaystyle \sin\left(2\sec^{-1}\left(\frac{u}{10}\right)\right)=\frac{20\sqrt{u^2-100}}{u^2}\text{ where } u>0$

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The answer is :
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If you have 20 oranges and a friend bought 10 more and gave then to you and you bought 10 more and then you divide the oranges f

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A diameter of a circle has enpoints (-2, 10) and (6, -4) in the standard (x, y) coordinate plane. What is the center of the circ

wlad13 [49]

Answer:

The center of the circle is $C(x,y) = (2, 3)$ .

Step-by-step explanation:

The center of the circle is the midpoint of the segment between the endpoints. We can determine the location of the center by this vectorial expression:

$C(x,y) = \frac{1}{2}\cdot R_{1}(x,y)+ \frac{1}{2}\cdot R_{2}(x,y)$ (1)

Where:

$C(x,y)$ - Center.

$R_{1} (x,y)$ , $R_{2} (x,y)$ - Location of the endpoints.

If we know that $R_{1} (x,y) = (-2,10)$ and $R_{2} (x,y) = (6,-4)$ , then the location of the center of the circle is:

$C(x,y) = \frac{1}{2}\cdot (-2,10)+\frac{1}{2}\cdot (6,-4)$

$C(x,y) = (-1, 5) + (3, -2)$

$C(x,y) = (2, 3)$

The center of the circle is $C(x,y) = (2, 3)$ .

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