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makkiz [27]
3 years ago
12

A teacher used the change of base formula to determine whether the equation below is correct.

Mathematics
1 answer:
Firlakuza [10]3 years ago
5 0

Answer:

Answer is B

The equation is correct since(log2^10)(log4^8)(log10^4)=log10/log2*log8/log4*log4/log10

=log8/log2

=3

Step-by-step explanation:

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The length of the other base is 82
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Factor 2x 2 - 10x - 12.
lana [24]
2(x+1)(x- 6) this is probably correct
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A machine can pack 80 cans in 5 minutes. Choose all of the values that show the ratio of cans to minutes.​
Drupady [299]

Answer:

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Step-by-step explanation:

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4 0
2 years ago
I thought of a number, added 4 5/7 to it, and got the number 12 times as big as my original what was my number
Genrish500 [490]
Let's call the number you thought of n. Then what the two steps you took can be written as an equation:
n+4\frac{5}{7}=12n

Subtract n to get all of your variables to one side:
4\frac{5}{7}=11n

At this point, I recommend turning your mixed number into an improper fraction. It will make things easier later on:
\frac{33}{7}=11n

Now divide both sides by 11 to get the value of n:
\frac{3}{7}=n
7 0
3 years ago
Read 2 more answers
How to prove this???
swat32
\cos^3 2A + 3 \cos 2A \\
\Rightarrow \cos 2A (\cos^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (\cos^2 2A + 3)  \\
\Rightarrow (\cos^2 A - \sin^2 A) (1 - \sin^2 2A + 3) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - \sin^2 2A) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - (2\sin A \cos A)(2\sin A \cos A)) \\
\Rightarrow (\cos^2 A - \sin^2 A) (4 - 4\sin^2 A \cos^2 A) \\ 
\Rightarrow 4(\cos^2 A - \sin^2 A) (1 - \sin^2 A \cos^2 A) 


go to right side now

4( \cos^6 A - \sin^6 A)\\
\Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A)

use x^3 - y^3 = (x-y)(x^2 + xy + y^2) and x^3 + y^3 = x^2 - xy + y^2

4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4( \cos^3 A - \sin^3 A)(\cos^3 A + \sin^3 A) \\
\Rightarrow  4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\
~\quad  \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A)

so \sin^2 A + \cos^2 A = 1

4( \cos^6 A - \sin^6 A)\\ \Rightarrow 4(\cos A - \sin A)(\cos^2 A + \cos A \sin A + \sin^2 A) \\ ~\quad \quad\cdot ( \cos A + \sin A)(\cos^2 A - \cos A \sin A + \cos^2 A) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 + \cos A \sin A )(1- \cos A \sin A ) \\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \cos^2 A \sin^2 A )\\ \Rightarrow 4(\cos^2 A - \sin^2 A)(1 - \sin^2 A \cos^2 A ) \\
 \Rightarrow Left hand side
4 0
3 years ago
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