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3 years ago

Write the prime factorization of the number 45

2 answers:

7 0

The prime factorization of a number means to factor out the number into multiples of primes.
Let's factor out 45.

45: 1 x 45, 3 x 15, 5 x 9.

Now we have 2 pairs of factors that can be broken down; 3&15, and 5&9.
Let's use 5 and 9.
9 can be broken down into 3 x 3.
We now have 5 x 3 x 3.
This is our factors into primes.

Your prime factorization of 45 is:
3 x 3 x 5.

I hope this helps!

Gnom [1K]3 years ago

7 0

5 times 3 times 3 is the correct answer or 5*3*3

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8mph

Step-by-step explanation:

when dina arrived to 20 masha was at 16. so the relation is 20/16 and if dina's speed was 10 your formula is 10/x=20/16 that means x/10=16/20. so x will be 8

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4 years ago

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3 years ago

a wildlife photographer spent 5 minutes taking pictures of a bison at a park. when the bison then decided she didn't want her ph

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The percent of time the photographer spent negotiating with the bison greater than the time he spent taking pictures is 500%

<h3>Percentage</h3>

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2 years ago

(5x+7)(8x+7) = 0<br>Find the two possible values.

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Prove or disprove (from i=0 to n) sum([2i]^4) <= (4n)^4. If true use induction, else give the smallest value of n that it doe

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Answer:

The statement is true for every n between 0 and 77 and it is false for $n\geq 78$

Step-by-step explanation:

First, observe that, for n=0 and n=1 the statement is true:

For n=0: $\sum^{n}_{i=0} (2i)^4=0 \leq 0=(4n)^4$

For n=1: $\sum^{n}_{i=0} (2i)^4=16 \leq 256=(4n)^4$

From this point we will assume that $n\geq 2$

As we can see, $\sum^{n}_{i=0} (2i)^4=\sum^{n}_{i=0} 16i^4=16\sum^{n}_{i=0} i^4$ and $(4n)^4=256n^4$ . Then,

$\sum^{n}_{i=0} (2i)^4 \leq(4n)^4 \iff \sum^{n}_{i=0} i^4 \leq 16n^4$

Now, we will use the formula for the sum of the first 4th powers:

$\sum^{n}_{i=0} i^4=\frac{n^5}{5} +\frac{n^4}{2} +\frac{n^3}{3}-\frac{n}{30}=\frac{6n^5+15n^4+10n^3-n}{30}$

Therefore:

$\sum^{n}_{i=0} i^4 \leq 16n^4 \iff \frac{6n^5+15n^4+10n^3-n}{30} \leq 16n^4 \\\\ \iff 6n^5+10n^3-n \leq 465n^4 \iff 465n^4-6n^5-10n^3+n\geq 0$

and, because $n \geq 0$ ,

$465n^4-6n^5-10n^3+n\geq 0 \iff n(465n^3-6n^4-10n^2+1)\geq 0 \\\iff 465n^3-6n^4-10n^2+1\geq 0 \iff 465n^3-6n^4-10n^2\geq -1\\\iff n^2(465n-6n^2-10)\geq -1$