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Tcecarenko [31]
3 years ago
14

Cot? 2x + cos2x + sin’ 2x = csc? 2x Need help with 40

Mathematics
1 answer:
kkurt [141]3 years ago
4 0

Answer:

\rule{100mm}{0.1mm}

    \cot^2(2x)+\cos^2(2x)+\sin^2(2x)=\csc^2(2x)

\dfrac{\cos^2(2x)}{\sin^2(2x)} +[(\cos^2(2x)+\sin^2(2x)]=\csc^2(2x)

                               \dfrac{\cos^2(2x)}{\sin^2(2x)} +1=\csc^2(2x)

                    \dfrac{\cos^2(2x)+\sin^2(2x)}{\sin^2(2x)}=\csc^2(2x)

                                    \dfrac{1}{sin^2(2x)} =\csc^2(2x)

                                     \boxed{\csc^2(2x)=\csc^2(2x)} ✔

\rule{100mm}{0.1mm}

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Answer:

$2 each

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5×2=$10 $40-$10=$30

Hope this helped! :)

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3 years ago
I have no idea how to do this! Can somebody please help me out a little?
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4 0
3 years ago
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9 &lt; 4(x+9) -7 - 25<br> Solve , intervals, and graph
Zigmanuir [339]

Answer:

x> 5/4

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9<4x+4

Step 2: Flip the equation.

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Step 3: Subtract 4 from both sides.

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pls give brainliest

5 0
3 years ago
What are all of the real roots of the following polynomial? f(x) x^4-24x^2-25
Greeley [361]

<u>Answer:</u>

x = ±5

<u>Step-by-step explanation:</u>

We are given the following polynomial function and we are to find all of its real roots:

x^4-24x^2-25

Let y=x^2 so we can now write it as:

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Factorizing it to get:

(y^2+y)+(-25y-25)

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Substitute back y=x^2 to get:

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\left ( x^2 + 1 \right ) \left ( x + 5 \right ) \left ( x - 5 \right )

The quadratic factor has only complex roots. Therefore, the real roots are x = ±5.

3 0
3 years ago
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