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2 years ago

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A coach is buying snacks for 22 players on a soccer match chip is a total of $77 to buy each player a bottle of water and energy

bar the price of one energy bar is two dollars

1 answer:

KIM [24]2 years ago

7 0

The question is incomplete:

A coach is buying snacks for 22 players at a soccer match. She pays a total of $77 to buy each player a bottle of water and an energy bar. The price of one energy bar is $2. Let w equal the price of a bottle of water. Write an equation that represents the situation.

Answer:

x=44+22w, where

x is the total amount paid

w is the price of a bottle of water

Step-by-step explanation:

With the information provided, you can say that the total amount paid is equal to the result of multiplying the price per energy bar for the number of energy bars purchased plus the result of multiplying the price per bottle of water for the number of bottles of water:

price per energy bar= $2

number of enrgy bars= 22

price per bottle of water= w

number of bottles of water= 22

x=(2*22)+(w*22)

x=44+22w, where

x is the total amount paid

w is the price of a bottle of water

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The value of x in the triangle is (a) 22

<h3>How to solve for x?</h3>

The complete question is in the attached image.

From the attached image of the triangle, we have:

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Evaluate sin(45)

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Divide

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2 years ago

Find the remaining trigonometric ratios of θ if csc(θ) = -6 and cos(θ) is positive

VikaD [51]

Now, the cosecant of θ is -6, or namely -6/1.

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we know the cosine is positive, and we know the opposite side is -1, or negative, the only happens in the IV quadrant, so θ is in the IV quadrant, now

$\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
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recall that

$\bf sin(\theta)=\cfrac{opposite}{hypotenuse}
\qquad\qquad 
cos(\theta)=\cfrac{adjacent}{hypotenuse}
\\\\\\
% tangent
tan(\theta)=\cfrac{opposite}{adjacent}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{adjacent}{opposite}
\\\\\\
% cosecant
csc(\theta)=\cfrac{hypotenuse}{opposite}
\qquad \qquad 
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sec(\theta)=\cfrac{hypotenuse}{adjacent}$

therefore, let's just plug that on the remaining ones,

$\bf sin(\theta)=\cfrac{-1}{6}
\qquad\qquad 
cos(\theta)=\cfrac{\sqrt{35}}{6}
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tan(\theta)=\cfrac{-1}{\sqrt{35}}
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% cotangent
cot(\theta)=\cfrac{\sqrt{35}}{1}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}$

now, let's rationalize the denominator on tangent and secant,

$\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
\\\\\\
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ollegr [7]

Answer:

$FC=25.9$

Step-by-step explanation:

<u>Congruent Triangles</u>

The figure shows two triangles FCD and FED. One of their sides is the radius of the circle, they both form a 90° angle with the line CD and ED and they share the same segment FD.

Thus, both triangles are congruent and we have the length of a side (12 units), one common side (the hypotenuse) and one unknown leg that must be equal. Therefore

$13x-16=4x+11$

Solving

$9x=27$

$x=3$

The required side FD is the hypotenuse of any of the triangles. The unknown leg can be calculated by

$13x-16=23$

Or

$4x+11=23$

Thus

$FD^2=12^2+23^2=673$

Solving

$FD=\sqrt{673}$

$\boxed{FC=25.9}$

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