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pentagon [3]
3 years ago
9

Lets be friends but you gots to match my energy (pwetty pwease; :) )

Mathematics
2 answers:
kondaur [170]3 years ago
7 0

Hi:) lets be friends hru I'm up for a conversation

jeka943 years ago
4 0
Give me points but sure
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Suppose theta is an angle in the standard position whose terminal side is in quadrant 4 and cot theta = -6/7. find the exact val
zimovet [89]

First off, let's notice that the angle is in the IV Quadrant, where sine is negative and the cosine is positive, likewise the opposite and adjacent angles respectively.

Also let's bear in mind that the hypotenuse is never negative, since it's simply just a radius unit.

\bf cot(\theta )=\cfrac{\stackrel{adjacent}{6}}{\stackrel{opposite}{-7}}\qquad \impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ c=\sqrt{6^2+(-7)^2}\implies c=\sqrt{36+49}\implies c=\sqrt{85} \\\\[-0.35em] ~\dotfill

\bf tan(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{adjacent}{6}} ~\hfill csc(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{opposite}{-7}} ~\hfill sec(\theta)=\cfrac{\stackrel{hypotenuse}{\sqrt{85}}}{\stackrel{adjacent}{6}} \\\\\\ sin(\theta)=\cfrac{\stackrel{opposite}{-7}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{sin(\theta)=\cfrac{-7}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies sin(\theta)=-\cfrac{7\sqrt{85}}{85}}

\bf cos(\theta)=\cfrac{\stackrel{adjacent}{6}}{\stackrel{hypotenuse}{\sqrt{85}}}\implies \stackrel{\textit{and rationalizing the denominator}}{cos(\theta)=\cfrac{6}{\sqrt{85}}\cdot \cfrac{\sqrt{85}}{\sqrt{85}}\implies cos(\theta)=\cfrac{6\sqrt{85}}{85}}

6 0
3 years ago
Do two 2 inch tiles add up to one 4 inch tile?
nikklg [1K]
Yes- I’m pretty sure
6 0
3 years ago
Read 2 more answers
I need help in these questions
Salsk061 [2.6K]

Answer:

see explanation

Step-by-step explanation:

All of these questions use the external angle theorem, that is

The external angle of a triangle is equal to the sum of the 2 opposite interior angles.

18

∠3 = 43° + 22° = 65°

19

∠2 + 71 = 92 ( subtract 71 from both sides )

∠2 = 21°

20

90 + ∠4 = 123 ( subtract 90 from both sides )

∠4 = 33°

21

2x - 15 + x - 5 = 148

3x - 20 = 148 ( add 20 to both sides )

3x = 168 ( divide both sides by 3 )

x = 56

Hence ∠ABC = x - 5 = 56 - 5 = 51°

22

2x + 27 + 2x - 11 = 100

4x + 16 = 100 ( subtract 16 from both sides )

4x = 84 ( divide both sides by 4 )

x = 21

Hence ∠JKL = 2x - 11 = (2 × 21) - 11 = 42 - 11 = 31°

3 0
4 years ago
Today, 32 customers at Tasha's Juice Emporium bought a total of 512 ounces of juice. How much juice did each customer buy on ave
Arturiano [62]
\frac{512 oz}{32 customers} = 16 ounces per customer
7 0
3 years ago
Read 2 more answers
In which quadrant of the coordinate graph does point R lie?
Sergio [31]
Your answer is d Its on the fourth quadrant! that Is really easy!
5 0
3 years ago
Read 2 more answers
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