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Zina [86]
2 years ago
6

Please help me on 4 and 5

Mathematics
2 answers:
babunello [35]2 years ago
8 0

Answer:

there is nothing there

Step-by-step explanation:

Aleksandr [31]2 years ago
3 0

Answer:

what's the question ?

.

.

.

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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
GIVING BRAINLIEST IF YOU DO IT CORRECT
abruzzese [7]

We use the word argument to refer to a series of reasons given to support a claim. The claim being supported is the conclusion. The reasons given to accept the conclusion are called premises. Analyzing an argument means identifying its premises and conclusion.

Hope this helps! God bless you, have a good day~!

8 0
3 years ago
Read 2 more answers
Two opposite sides of a parallelogram are 3x-8 and 10x+13. Find the measure of all sides
RSB [31]

Answer:

Step-by-step explanation:

Using the parallelogram theorem

3x-8 = 10x+13

-7x = 21

x = -3

Substitute

3(-3) - 8

-17

10(-3) + 13

-17

5 0
3 years ago
Marco and Drew stacked boxes on a shelf. Marco lifted 9 boxes and Drew lifted 14 boxes. The boxes that Drew lifted each weighed
MrRa [10]
The correct answer is b. 14(m-8).

This is because it says that the weight of the boxes drew lifted were 8 pounds less than what marco lifted.  This can be represented by m-8.  You then need to multiply this by the amount of boxes that drew lifted, or 14.  
8 0
3 years ago
Find the equation of a line that is parallel to the line y=-3x-1 and crosses the y-axis at (0,-8).
Colt1911 [192]

Answer:

Y=-3x-3

Step-by-step explanation:

3 0
2 years ago
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