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SSSSS [86.1K]
1 year ago
12

3x-1=4x+8-x Find X Value

Mathematics
2 answers:
Marrrta [24]1 year ago
6 0

Answer:

is the question correct, you end up with no X's

3x -4x + x = 9

or I've got it wrong

ss7ja [257]1 year ago
5 0
X just disappears, it doesn’t have a value
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5 0
3 years ago
First question, thanks. I believe there should be 3 answers
zysi [14]

Given: The following functions

A)cos^2\theta=sin^2\theta-1B)sin\theta=\frac{1}{csc\theta}\begin{gathered} C)sec\theta=\frac{1}{cot\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}

To Determine: The trigonometry identities given in the functions

Solution

Verify each of the given function

\begin{gathered} cos^2\theta=sin^2\theta-1 \\ Note\text{ that} \\ sin^2\theta+cos^2\theta=1 \\ cos^2\theta=1-sin^2\theta \\ Therefore \\ cos^2\theta sin^2\theta-1,NOT\text{ }IDENTITIES \end{gathered}

B

\begin{gathered} sin\theta=\frac{1}{csc\theta} \\ Note\text{ that} \\ csc\theta=\frac{1}{sin\theta} \\ sin\theta\times csc\theta=1 \\ sin\theta=\frac{1}{csc\theta} \\ Therefore \\ sin\theta=\frac{1}{csc\theta},is\text{ an identities} \end{gathered}

C

\begin{gathered} sec\theta=\frac{1}{cot\theta} \\ note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ tan\theta cot\theta=1 \\ tan\theta=\frac{1}{cot\theta} \\ Therefore, \\ sec\theta\ne\frac{1}{cot\theta},NOT\text{ IDENTITY} \end{gathered}

D

\begin{gathered} cot\theta=\frac{cos\theta}{sin\theta} \\ Note\text{ that} \\ cot\theta=\frac{1}{tan\theta} \\ cot\theta=1\div tan\theta \\ tan\theta=\frac{sin\theta}{cos\theta} \\ So, \\ cot\theta=1\div\frac{sin\theta}{cos\theta} \\ cot\theta=1\times\frac{cos\theta}{sin\theta} \\ cot\theta=\frac{cos\theta}{sin\theta} \\ Therefore \\ cot\theta=\frac{cos\theta}{sin\theta},is\text{ an Identity} \end{gathered}

E

\begin{gathered} 1+cot^2\theta=csc^2\theta \\ csc^2\theta-cot^2\theta=1 \\ csc^2\theta=\frac{1}{sin^2\theta} \\ cot^2\theta=\frac{cos^2\theta}{sin^2\theta} \\ So, \\ \frac{1}{sin^2\theta}-\frac{cos^2\theta}{sin^2\theta} \\ \frac{1-cos^2\theta}{sin^2\theta} \\ Note, \\ cos^2\theta+sin^2\theta=1 \\ sin^2\theta=1-cos^2\theta \\ So, \\ \frac{1-cos^2\theta}{sin^2\theta}=\frac{sin^2\theta}{sin^2\theta}=1 \\ Therefore \\ 1+cot^2\theta=csc^2\theta,\text{ is an Identity} \end{gathered}

Hence, the following are identities

\begin{gathered} B)sin\theta=\frac{1}{csc\theta} \\ D)cot\theta=\frac{cos\theta}{sin\theta} \\ E)1+cot^2\theta=csc^2\theta \end{gathered}

The marked are the trigonometric identities

3 0
1 year ago
Y= -2 (x+2)^2-8<br>vertex=?<br>how to graph?
MissTica
Https://www.desmos.com/calculator you can use this calculator just plug the what y= in it and it will show you the graph and the vertex
 
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3 years ago
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The radius r of a circle can be written as a function of the area A with the following equation: r= sqrt(A/pi) What is the domai
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The domain of the function is \[A \ge0\] If A is negative we get an imaginary number for the radius. This makes sense because the area of a circle cannot be negative.
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Find the following System by the cramer's rule 4x-2y=8 3x+y=-4
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Answer:

the value of x is 0 and y is -4.

hope it helps uh..

4 0
3 years ago
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