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2 years ago

PLS HELP SOLVE FOR X

Mathematics

2 answers:

horsena [70]2 years ago

8 0

Answer:

x=26

Step-by-step explanation:

AleksandrR [38]2 years ago

7 0

Answer:

X=26

Step-by-step explanation:

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Use the midpoint formula to estimate the sales of Cars, Inc. in 2009, given the sales in 2008 and 2010. Assume that the sales of

lana [24]

Answer:

<em>The estimated sales were $260 million</em>

Step-by-step explanation:

Assume the endpoints of a segment are (x1,y1) and (x2,y2).

The midpoint (xm,ym) is calculated as follows:

$\displaystyle x_m=\frac{x_1+x_2}{2}$

$\displaystyle y_m=\frac{y_1+y_2}{2}$

The sales ov Cars, Inc. were (2008,240 million) and (2010,280 million). We need to use the midpoint to estimate the sales in 2009:

$\displaystyle x_m=\frac{2008+2010}{2}=2009$

$\displaystyle y_m=\frac{240+280}{2}=260$

The estimated sales were $260 million

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2 years ago

1.) A tower casts a shadow that is 60 feet long when the angle

Fynjy0 [20]

Answer:

128.67 ft

Step-by-step explanation:

Let the height of the tower be x feet.

$\tan \: 65 \degree = \frac{x}{60} \\ \\ 2.1445069205 = \frac{x}{60} \\ \\ x = 2.1445069205 \times 60 \\ \\ x = 128.670415 \\ \\ x = 128.67 \: ft$

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2 years ago

The movie theater has 250 seats. 225 seats were sold for the current showing. What percent of seats are empty?

disa [49]

Answer:

Step-by-step explanation:

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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.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

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

$\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}
\\\\\\
\textit{using the pythagorean theorem}\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a
\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
\end{cases}
\\\\\\
\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}$

recall that

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

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

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

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

$\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}$

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3 years ago

Which amount is bigger

erastova [34]

What the two or more numbers or amounts? Something needs to be compared

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