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ludmilkaskok [199]
2 years ago
11

Which algebraic expression is equivalent to this expression?

Mathematics
1 answer:
Leno4ka [110]2 years ago
3 0

The numerical value of this algebraic expression is 123

  • Step-by-step explanation:

To solve this algebraic expression, we're going to subtract the numbers indicated in parentheses, and finally, we're going to multiply and <em>add </em>the numbers.

  • Resolution:

\large \sf =6(22 - 12) + 63

\large \sf =6(10) + 63

\large \sf =60 + 63

\boxed{\boxed{{\large \sf 123}}}

Therefore, we can conclude that the numeric value of this expression will be 123

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Which of these functions has an inverse function? Select all that apply.
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Simplify: cos2x-cos4 all over sin2x + sin 4x
GrogVix [38]

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\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}=\tan\left(x\right)

Step-by-step explanation:

\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}

Apply formula:

\cos\left(A\right)-\cos\left(B\right)=-2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right) and

\sin\left(A\right)+\sin\left(B\right)=2\cdot\sin\left(\frac{A+B}{2}\right)\cdot\sin\left(\frac{A-B}{2}\right)

We get:

=\frac{-2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\sin\left(\frac{2x-4x}{2}\right)}{2\cdot\sin\left(\frac{2x+4x}{2}\right)\cdot\cos\left(\frac{2x-4x}{2}\right)}

=\frac{-\sin\left(\frac{2x-4x}{2}\right)}{\cos\left(\frac{2x-4x}{2}\right)}

=\frac{-\sin\left(\frac{-2x}{2}\right)}{\cos\left(\frac{-2x}{2}\right)}

=\frac{-\sin\left(-x\right)}{\cos\left(-x\right)}

=\frac{-\cdot-\sin\left(x\right)}{\cos\left(x\right)}

=\frac{\sin\left(x\right)}{\cos\left(x\right)}

=\tan\left(x\right)

Hence final answer is

\frac{\cos\left(2x\right)-\cos\left(4x\right)}{\sin\left(2x\right)+\sin\left(4x\right)}=\tan\left(x\right)

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