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3 years ago

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

Step-by-step explanation:

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3 years ago

Factor 2x 2 - 10x - 12.

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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.

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

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

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3 years ago