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4 years ago

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

boyakko [2]

The probability is 1/5

7 0

3 years ago

Please help me I suck at math

MatroZZZ [7]

Answer:

Step-by-step explanation:

Find the slope.

Then the y int is 1

It also says less than or equal to

so you shade under

4 0

3 years ago

Find X X+3 5<br> —— = —<br> 8 4<br>

ExtremeBDS [4]

Exact Form:
x
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Decimal Form:
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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

#SPJ1

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