6. Write in exponential form: log(3)81 = 4

Mathematics

1 answer:

earnstyle [38]3 years ago

4 0

Answer:

$81 = {3}^{4}$

Step-by-step explanation:

$log_{3}(81) = 4 \\ \\ 81 = {3}^{4}$

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Evaluate the expression 5(8y+3r-5z) if r=3,y=5/8,and z=2

snow_lady [41]

First you plug in the missing values
5(8×5/8+3×3-5×2) = 20

3 0

3 years ago

12n^3 + 16n^3 can someone please help me with this?

vodka [1.7K]

Answer:

$4n^{3} (3n^{2} + 4)$

Step-by-step explanation:

Assuming the problem is asking you to factor out the formula, it can be done easily by finding a common number with the two givens.

4 goes into both 12 and 16 evenly, so we will use this to factor. But that's not all, we have the n-variable to worry about.

Look at the n-variable exponents in the problem, take the highest power that can go be subtracted from both and use that. We have a $n^{5}$ and $n^{3}$ . Since 5 can't go into 3, we must use 3 to factor.

Our factoring variable will therefore be $4n^{3}$ . Now, use this to simply divide and get the answer:

$( 12n^{5} + 16n^{3} ) / 4n^{3} = 4n^{3}(3n^{2} + 4)$

So how did we get this? For the whole numbers, we have to divide: (12/4, 16/4). Next, we have to minus the exponents since we are dividing.

When dividing the 12 and 4, minus the $n^{5}$ and $n^{3}$ to get $n^{2}$ . Get rid of the $n^{3}$ term for the 16 and leave the 4. And there's your answer!