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MAXImum [283]
4 years ago
14

How many times smaller is 3 × 10-7 than 6 × 10-6?

Mathematics
1 answer:
Luba_88 [7]4 years ago
6 0
3*10-7=23
6*10-6=54
it is 2.35 times smaller. 
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Illene wrapped 5 gifts for her 5 cousins. if she gives them out randomly, what is the probability that every cousin gets the rig
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The probability is 1/5
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3 years ago
Please help me I suck at math ​
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D.

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Then the y int is 1

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3 years ago
Find X X+3 5<br> —— = —<br> 8 4<br>​
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4 0
4 years ago
Rational expressions are often used in combining rates of work.
denis-greek [22]

Rational expressions are often used in combining rates of work. Therefore, it's true.

<h3>What is rational expression?</h3>

It should be noted that a rational expression is simply defined by a rational fraction.

They're are used in combining rates of work. Fir example, if Mr John performs 1/2 of his work and does 1/3 on another day. This can be expressed as:

= 1/2 + 1/3

= 5/6

Learn more about rational expression on:

brainly.com/question/2264785

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7 0
2 years ago
Use mathematical induction to prove the statement is true for all positive integers n. 1^2 + 3^2 + 5^2 + ... + (2n-1)^2 = (n(2n-
Charra [1.4K]

Answer:

The statement is true is for any n\in \mathbb{N}.

Step-by-step explanation:

First, we check the identity for n = 1:

(2\cdot 1 - 1)^{2} = \frac{2\cdot (2\cdot 1 - 1)\cdot (2\cdot 1 + 1)}{3}

1 = \frac{1\cdot 1\cdot 3}{3}

1 = 1

The statement is true for n = 1.

Then, we have to check that identity is true for n = k+1, under the assumption that n = k is true:

(1^{2}+2^{2}+3^{2}+...+k^{2}) + [2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}

\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)}{3} +[2\cdot (k+1)-1]^{2} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}

\frac{k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot [2\cdot (k+1)-1]^{2}}{3} = \frac{(k+1)\cdot [2\cdot (k+1)-1]\cdot [2\cdot (k+1)+1]}{3}

k\cdot (2\cdot k -1)\cdot (2\cdot k +1)+3\cdot (2\cdot k +1)^{2} = (k+1)\cdot (2\cdot k +1)\cdot (2\cdot k +3)

(2\cdot k +1)\cdot [k\cdot (2\cdot k -1)+3\cdot (2\cdot k +1)] = (k+1) \cdot (2\cdot k +1)\cdot (2\cdot k +3)

k\cdot (2\cdot k - 1)+3\cdot (2\cdot k +1) = (k + 1)\cdot (2\cdot k +3)

2\cdot k^{2}+5\cdot k +3 = (k+1)\cdot (2\cdot k + 3)

(k+1)\cdot (2\cdot k + 3) = (k+1)\cdot (2\cdot k + 3)

Therefore, the statement is true for any n\in \mathbb{N}.

4 0
3 years ago
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