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kolbaska11 [484]
3 years ago
7

Can someone please help me with this whole page asap !!!

Mathematics
1 answer:
Valentin [98]3 years ago
4 0
What answers do u need
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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
4. Find the solution to each equation mentally.
White raven [17]
A, 10. b,-300. c, -1. d, -3567.
youre basically looking for the number that gets you to the other number. how far away from 30 is 40? 10. how far is 500 from 200? 300, but in the other direction so its negative. and for D since the final number is 0 just use the same number but flip the sign
5 0
3 years ago
Read 2 more answers
How many solutions do you think the equation -10x+3=5x-27 has? Explain. Then, find the solution(s).
deff fn [24]
This will only have one solution because it does not deal with a quadratic. Then to isolate x you need to get the similar terms on different sides by subtracting or adding, Then divide both sides by the number next to x (coefficient).
4 0
3 years ago
C+-c=0<br> Identity Property<br> Associative Property<br> Communtative Property<br> Inverse Property
Zanzabum

Answer:

⣠⣴⣶⣿⠿⢿⣶⣶⣦⣄⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⣼⡿⠋⠁⠀⠀⠀⢀⣈⠙⢿⣷⡄⠀⠀ ⠀⠀⠀⠀⢸⣿⠁⠀⢀⣴⣿⠿⠿⠿⠿⠿⢿⣷⣄⠀ ⠀⢀⣀⣠⣾⣿⡇⠀⣾⣿⡄⠀⠀⠀⠀⠀⠀⠀⠹⣧ ⣾⡿⠉⠉⣿⠀⡇⠀⠸⣿⡌⠓⠶⠤⣤⡤⠶⢚⣻⡟ . ⣿⣧⠖⠒⣿⡄⡇⠀⠀⠙⢿⣷⣶⣶⣶⣶⣶⢿⣿⠀ . ⣿⡇⠀⠀⣿⡇⢰⠀⠀⠀⠀⠈⠉⠉⠉⠁⠀⠀⣿⠀. ⣿⡇⠀⠀⣿⡇⠈⡄⠀⠀⠀⠀⠀⠀⠀⠀⢀⣿⣿⠀ ⣿⣷⠀⠀⣿⡇⠀⠘⠦⣄⣀⣀⣀⣀⣀⡤⠊⠀⣿⠀ ⢿⣿⣤⣀⣿⡇⠀⠀⠀⢀⣀⣉⡉⠁⣀⡀⠀⣾⡟⠀ ⠀⠉⠛⠛⣿⡇⠀⠀⠀⠀⣿⡟⣿⡟⠋⠀ * ° * • ☆ ° .°• * ✯ ☄ ☆ ★ * ° * °· * . • ° ★ • ☄ ☄ ▁▂▃▄▅▆▇▇▆▅▄▃▁

⣠⣴⣶⣿⠿⢿⣶⣶⣦⣄⠀⠀⠀⠀ ⠀⠀⠀⠀⠀⣼⡿⠋⠁⠀⠀⠀⢀⣈⠙⢿⣷⡄⠀⠀ ⠀⠀⠀⠀⢸⣿⠁⠀⢀⣴⣿⠿⠿⠿⠿⠿⢿⣷⣄⠀ ⠀⢀⣀⣠⣾⣿⡇⠀⣾⣿⡄⠀⠀⠀⠀⠀⠀⠀⠹⣧ ⣾⡿⠉⠉⣿⠀⡇⠀⠸⣿⡌⠓⠶⠤⣤⡤⠶⢚⣻⡟ . ⣿⣧⠖⠒⣿⡄⡇⠀⠀⠙⢿⣷⣶⣶⣶⣶⣶⢿⣿⠀ . ⣿⡇⠀⠀⣿⡇⢰⠀⠀⠀⠀⠈⠉⠉⠉⠁⠀⠀⣿⠀. ⣿⡇⠀⠀⣿⡇⠈⡄⠀⠀⠀⠀⠀⠀⠀⠀⢀⣿⣿⠀ ⣿⣷⠀⠀⣿⡇⠀⠘⠦⣄⣀⣀⣀⣀⣀⡤⠊⠀⣿⠀ ⢿⣿⣤⣀⣿⡇⠀⠀⠀⢀⣀⣉⡉⠁⣀⡀⠀⣾⡟⠀ ⠀⠉⠛⠛⣿⡇⠀⠀⠀⠀⣿⡟⣿⡟⠋⠀ * ° * • ☆ ° .°• * ✯ ☄ ☆ ★ * ° * °· * . • ° ★ • ☄ ☄ ▁▂▃▄▅▆▇▇▆▅▄▃▁

7 0
2 years ago
Find the indicated derivative If f(x) = x^3 , find f’(-4)
babunello [35]

Answer:

48

Step-by-step explanation:

f(x) = x³

f'(x) = 3x²

f'(-4) = 3(-4)²

f'(-4) = 48

7 0
2 years ago
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