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KATRIN_1 [288]
2 years ago
10

PLZ HELP ME WILL GIVE BRAINLIEST

Mathematics
1 answer:
nadezda [96]2 years ago
3 0

Answer:

1.  2 \times 10^{- 4} = 2 \div 10^{4}

2. (- 10)^{2} = 100 and - 10^{2} = - 100

3. (a.b.c. ........)^{n} = a^{n}.b^{n}.c^{n}. ..........

Example:  (30)^{4} = (2 \times 3 \times 5)^{4} = 2^{4} \times  3^{4} \times 5^{4}

Step-by-step explanation:

1. We have 2 \times 10^{- 4}, and we have to prove that this term is equivalent to 2 \div 10^{4}.

Now, 2 \times 10^{- 4}

= 2 \times \frac{1}{10^{4} }  

{Since we know the property of exponent as a^{- b} = \frac{1}{a^{b} } }

= \frac{2}{10^{4} }

= 2 \div 10^{4}

2. (- 10)^{2} = (- 10) \times (- 10) = 100 and  

- 10^{2} = - [(10) \times (10)] = - 100

3. The power of products property gives

(a.b.c. ........)^{n} = a^{n}.b^{n}.c^{n}. .......... ........ (1)

For example, we can write, (30)^{4} = (2 \times 3 \times 5)^{4} = 2^{4} \times  3^{4} \times 5^{4} = 810000

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The average of six positive integers starting with a is equal to b. What is the average of five consecutive integers ending with
kvasek [131]

The average of the five consecutive numbers ending with b in discuss when expressed in terms of a is; Choice D; a+3.

<h3>What is the average of five consecutive integers ending with b?</h3>

First, since it was given in the task content that the average of six positive consecutive odd integers starting with a is equal to b, it therefore follows that;

(a+a+2+a+4+a+6+a+8+a+10)/6 = b

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Also, let the average of the consecutive intergers ending with b be denoted by; x.

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Ultimately, the value of the required average is; = a+5-2 = a+3.

Read more on average of integers;

brainly.com/question/24369824

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Step-by-step explanation:

Thank me later

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