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2 years ago

What is the gcf of 18a and 20ab

Mathematics

2 answers:

oksian1 [2.3K]2 years ago

7 0

Answer:

Step-by-step explanation:

the biggest number they can both divide by is 2.

ohaa [14]2 years ago

4 0

Answer:

It is 2a

Step-by-step explanation:

Ok the greatest common factor is the number and/or term common to a seri3s of values so you separate them

the GCF of 18 and 20 is 2

and you also have a and ab

now, what is common to them both

so you would get the gcf as 2a so it would be written as 2a(9+10b)

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X(2x-5)+x(2x-3)=(4x+6)(x-3)

Helga [31]

Answer:

x=9

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

End of year assessment

tensa zangetsu [6.8K]

Answer:

Don't fail

Step-by-step explanation:

4 0

2 years ago

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

ddd [48]

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 $n \geq 0$ ,

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

Observe that, because $n \geq 2$ and is an integer,

$n^2(465n-6n^2-10)\geq -1 \iff 465n-6n^2-10 \geq 0 \iff n(465-6n) \geq 10\\\iff 465-6n \geq 0 \iff n \leq \frac{465}{6}=\frac{155}{2}=77.5$

In concusion, the statement is true if and only if n is a non negative integer such that $n\leq 77$

So, 78 is the smallest value of n that does not satisfy the inequality.

Note: If you compute $(4n)^4- \sum^{n}_{i=0} (2i)^4$ for 77 and 78 you will obtain:

$(4n)^4- \sum^{n}_{i=0} (2i)^4=53810064$

$(4n)^4- \sum^{n}_{i=0} (2i)^4=-61754992$

7 0

3 years ago

1. Find the Least Common Multiple of these two monomials: See picture

gtnhenbr [62]

Answer:

The last choice is correct

$LCM=120a^4b^7c^5$

Step-by-step explanation:

Least Common Multiple (LCM)

To find the LCM we can follow this procedure:

List the prime factors of each monomial.

Multiply each factor the greatest number of times it occurs in either factor.

We have two monomials:

$12a^4b^2c^5$

$40a^3b^7c^1$

The prime factors of the first monomial are:

$2^2,3,a^4,b^2,c^5$

The prime factors of the second monomial are:

$2^3,5,a^3b^7c^1$

LCM = Multiply $2^3*3*5*a^4*b^7*c^5$

These are all the factors the greatest number of times they occur.

Operating:

$LCM=8*15*a^4*b^7*c^5$

$\boxed{LCM=120a^4b^7c^5}$

The last choice is correct

3 0

3 years ago

URGENTS PLLLLLSSSSSSSSS

Arlecino [84]

Just divide 300 by 4 since 25% is 1/4
300/4 = 75
D. 75
Hope this helps!

5 0

3 years ago

Read 2 more answers