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Montano1993 [528]
3 years ago
5

Help me with that please

Mathematics
1 answer:
Sonbull [250]3 years ago
7 0
About $5.26 will be deducted from $50

So he will get $44.74
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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}
3 0
3 years ago
True or False?<br><br> 9.00 = 9 = 9.0
Damm [24]
The answer to this is True.
3 0
3 years ago
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PLS HELP ME WITH GEOMETRY PLS :)
MArishka [77]

Answer:

-4, -24

Step-by-step explanation:

The location of Z is either +10 or -10 units away from Y, since YZ = 10. Then Z could be located at either -4 or -24.

3 0
3 years ago
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Two parallel lines are crossed a transversal . What is the value of k
vredina [299]
They would be alternate exterior so they would have to to be equal
2k + 11 = 131
-11 -11
2k = 120
---- -----
2k 2k
k = 60
7 0
3 years ago
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Please help! Not even sure how to start with this one or how to calculate
vivado [14]

Remember when you learning about the "Least Common
Multiple" and you wondered when in the world you would
ever need to use it ?
Well, here you are !   Hello !

The next time they will cross the starting line together is the
LCM (Least Common Multiple) of 30 and 40 .

Do you remember how to find it ?
Here's one easy way: 

Count by 30s until you come to a number that's divisible by 40 .

Laura's first lap. . . . . . . . 30 seconds . . . not divisible by 40
Laura's two laps . . . . . . . 60 seconds . . . not divisible by 40
Laura's three laps. . . . . . 90 seconds . . . not divisible by 40
Laura's four laps. . . . . . 120 seconds . . . divisible by 40

After 120 seconds, Laura has completed four laps,
Shelby has completed only three laps.

4 0
3 years ago
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