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Tems11 [23]
2 years ago
8

Simplify the following. Express your answer with positive exponents

Mathematics
1 answer:
oksano4ka [1.4K]2 years ago
8 0

On simplification, the expression is 15b^10/a^5

Step-by-step explanation:

  • Step 1: Use the laws of exponents to solve this. a^m · a^n = a^m+n and a^m ÷ a^n = a^m-n.
  • Answer is shown in attachment.

                                                                 

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3%, 2/50, 0.08 is ur answer!!
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2 years ago
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If log2 5 = k, determine an expression for log32 5 in terms of k.
lukranit [14]

Answer:

log_3_2(5)=\frac{1}{5} k

Step-by-step explanation:

Let's start by using change of base property:

log_b(x)=\frac{log_a(x)}{log_a(b)}

So, for log_2(5)

log_2(5)=k=\frac{log(5)}{log(2)}\hspace{10}(1)

Now, using change of base for log_3_2(5)

log_3_2(5)=\frac{log(5)}{log(32)}

You can express 32 as:

2^5

Using reduction of power property:

log_z(x^y)=ylog_z(x)

log(32)=log(2^5)=5log(2)

Therefore:

log_3_2(5)=\frac{log(5)}{5*log(2)}=\frac{1}{5} \frac{log(5)}{log(2)}\hspace{10}(2)

As you can see the only difference between (1) and (2) is the coefficient \frac{1}{5} :

So:

\frac{log(5)}{log(2)} =k\\

log_3_2(5)=\frac{1}{5} \frac{log(5)}{log(2)} =\frac{1}{5} k

6 0
3 years ago
State the necessary value of nn that will make each statement true.<br> 0.0004 × 0.002 = 8 × 10^n
vodka [1.7K]

Answer:

-7 = n

Step-by-step explanation:

0,002 \times 0,0004 = 0,0000008

0,0000008 = 8 \times {10}^{-7}

* You must make sure that you have seven zeros in your decimal.

I am joyous to assist you anytime.

6 0
3 years ago
Compare 6.23x10^14 and 8.912x10^12
Tju [1.3M]

Answer:

hello : 6.23x10^14 >  8.912x10^12^12

Step-by-step explanation:

let : A= 6.23x10^14 and  B =  8.912x10^12

A= (6.23x10^2) x10^12  ... because : 10^14 = 10^2x10^12

A = 623 x 10^12

623 > 8.912

so : A > B

3 0
3 years ago
What is the length of AC?
Elan Coil [88]

well, the triangle is an isosceles, so it twin sides, and the twin sides make twin angles, as we see by the tickmarks on A and C, meaning AB = BC.

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