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IrinaVladis [17]
2 years ago
15

FREEEEE plz sub to my yt it called chrisgamez it the one with like 27-30 subs ty

Mathematics
2 answers:
Vesna [10]2 years ago
7 0
Yes ofc I will thanks for free points
Butoxors [25]2 years ago
3 0

Answer:

Ayo, you should join my discord server discord. gg/cnPhpUJF

Step-by-step explanation:

Thanks for the points

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Find the circumference of the circle below to the nearest tenth Use 3.14 for a
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Answer:

125.6

Step-by-step explanation:

all you have to do is multiple the 3.14 with 40 cm to get 125.6 cm

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2 years ago
Please help. If f(x)= 2^x-1+3 and g(x)=5x-9 what is (f-g)(x)??
Readme [11.4K]

( f + g ) (x) = –2x + 6

( f – g ) (x) = 8x – 2

( f  × g ) (x) = –15x2 + 2x + 8

<span>\mathbf{\color{purple}{ \left(\small{\dfrac{\mathit{f}}{\mathit{g}}}\right)(\mathit{x}) = \small{\dfrac{3\mathit{x} + 2}{4 - 5\mathit{x}}} }}<span><span>(<span><span>​g</span>​<span>​f</span><span>​​</span></span>)</span>(x)=<span><span>​<span>4−5x</span></span>​<span>​<span>3x+2</span></span><span>​​</span></span></span></span>

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2 years ago
Read 2 more answers
Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______
Ludmilka [50]

Answer:

(a)\ \sec^2(\theta) = 82

(b)\ \cot(\theta) = \frac{1}{9}

(c)\ \cot(\frac{\pi}{2} - \theta) = 9

(d)\ \csc^2(\theta) = \frac{82}{81}

Step-by-step explanation:

Given

\tan(\theta) = 9

Required

Solve (a) to (d)

Using tan formula, we have:

\tan(\theta) = \frac{Opposite}{Adjacent}

This gives:

\frac{Opposite}{Adjacent} = 9

Rewrite as:

\frac{Opposite}{Adjacent} = \frac{9}{1}

Using a unit ratio;

Opposite = 9; Adjacent = 1

Using Pythagoras theorem, we have:

Hypotenuse^2 = Opposite^2 + Adjacent^2

Hypotenuse^2 = 9^2 + 1^2

Hypotenuse^2 = 81 + 1

Hypotenuse^2 = 82

Take square roots of both sides

Hypotenuse =\sqrt{82}

So, we have:

Opposite = 9; Adjacent = 1

Hypotenuse =\sqrt{82}

Solving (a):

\sec^2(\theta)

This is calculated as:

\sec^2(\theta) = (\sec(\theta))^2

\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2

Where:

\cos(\theta) = \frac{Adjacent}{Hypotenuse}

\cos(\theta) = \frac{1}{\sqrt{82}}

So:

\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2

\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2

\sec^2(\theta) = (\sqrt{82})^2

\sec^2(\theta) = 82

Solving (b):

\cot(\theta)

This is calculated as:

\cot(\theta) = \frac{1}{\tan(\theta)}

Where:

\tan(\theta) = 9 ---- given

So:

\cot(\theta) = \frac{1}{\tan(\theta)}

\cot(\theta) = \frac{1}{9}

Solving (c):

\cot(\frac{\pi}{2} - \theta)

In trigonometry:

\cot(\frac{\pi}{2} - \theta) = \tan(\theta)

Hence:

\cot(\frac{\pi}{2} - \theta) = 9

Solving (d):

\csc^2(\theta)

This is calculated as:

\csc^2(\theta) = (\csc(\theta))^2

\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2

Where:

\sin(\theta) = \frac{Opposite}{Hypotenuse}

\sin(\theta) = \frac{9}{\sqrt{82}}

So:

\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2

\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2

\csc^2(\theta) = \frac{82}{81}

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Use the rule x + 8 to write a pattern. Begin with x = 0.
Nutka1998 [239]
8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104
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What is 24.45379+5679.0000000
Scilla [17]
It equals <span>5703.45379.

Happy studying ^_^</span>
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