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3 years ago

What is the GCF of

Mathematics

2 answers:

Troyanec [42]3 years ago

7 0

$1.) 14b^2 - 35b^3=2\cdot7b^2-5b\cdot7b^2=7b^2(2-5b)\\\\2.) 18x^3y^2 + 42x^2y^3=3x\cdot6x^2y^2+7y\cdot6x^2y^2=6x^2y^2(3x+7y)\\\\3.) 12x^2 + 16y^2=4(3x^2+4y^2)\\\\4.) 3x^2 - 9xy + 12y=3(x^2-3xy+4)\\\\5.) 45x^3 - 60xy^2=3x^2\cdot15x-4y^2\cdot15x=15x(3x^2-4y^2)\\\\6.)36x^2+18x^3-6x+12x^4=6x(2x^3+3x^2+6x-1)$

$Ans.\ GFC:\ 1)\ 7b^2,\ 2)\ 6x^2y^2,\ 3)\ 4,\ 4)\ 3,\ 5)\ 15x,\ 6)\ 6x$

natima [27]3 years ago

6 0

$1.\\14b^2-35b^3\\GCF(14;35)=7\\GCF(b^2;b^3)=b^2\\14b^2-35b^3=7b^2(2-5b)\\\\2.\\18x^3y^2+42x^2y^3\\GCF(18;42)=6\\GCF(x^3y^2;x^2y^3)=x^2y^2\\18x^3y^2+42x^2y^3=6x^2y^2(3x+7y)$

$3.\\12x^2+16y^2\\GCF(12;16)=4\\GCF(x^2;y^2)=1\\12x^2+16y^2=4(3x^2+4y^2)\\\\4.\\3x^2-9xy+12y\\GCF(3;9;12)=3\\GCF(x^2;xy;y)=1\\3x^2-9xy+12y=3(x^2-3xy+4y)$

$5.\\45x^3-60xy^2\\GCF(45;60)=15\\GCF(x^3;xy^2)=x\\45x^3-60xy^2=15x(3x^2-4y^2)\\\\6.\\36x^2+18x^3-6x+12x^4\\GCF(36;18;6;12)=6\\GCF(x^2;x^3;x;x^4)=x\\36x^2+18x^3-6x+12x^4=6x(6x+3x^2-1+2x^3)$

$GCF:\\1)\ b^2;\ 2)\ 6x^2y^2;\ 3)\ 4;\ 4);\ 3;\ 5);\ 15x;\ 6)\ 6x$

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Determine formula of the nth term 2, 6, 12 20 30,42

nalin [4]

Check the forward differences of the sequence.

If $\{a_n\} = \{2,6,12,20,30,42,\ldots\}$ , then let $\{b_n\}$ be the sequence of first-order differences of $\{a_n\}$ . That is, for n ≥ 1,

$b_n = a_{n+1} - a_n$

so that $\{b_n\} = \{4, 6, 8, 10, 12, \ldots\}$ .

Let $\{c_n\}$ be the sequence of differences of $\{b_n\}$ ,

$c_n = b_{n+1} - b_n$

and we see that this is a constant sequence, $\{c_n\} = \{2, 2, 2, 2, \ldots\}$ . In other words, $\{b_n\}$ is an arithmetic sequence with common difference between terms of 2. That is,

$2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2$

and we can solve for $b_n$ in terms of $b_1=4$ :

$b_{n+1} = b_n + 2$

$b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2$

$b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2$

and so on down to

$b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)$

We solve for $a_n$ in the same way.

$2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)$

Then

$a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)$

$a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))$

$a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))$

and so on down to

$a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2$

$\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}$

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2 years ago

11. Keng creates a painting on a rectangular canvas with a width that is four inches longer

GrogVix [38]

Part A

<h3>Answer: h^2 + 4h</h3>

-------------------

Explanation:

We multiply the length and height to get the area

area = (length)*(height)

area = (h+4)*(h)

area = h(h+4)

area = h^2 + 4h .... apply the distributive property

The units for the area are in square inches.

===========================================================

Part B

<h3>Answer: h^2 + 16h + 60</h3>

-------------------

Explanation:

If we add a 3 inch frame along the border, then we're adding two copies of 3 inches along the bottom side. The h+4 along the bottom updates to h+4+3+3 = h+10 along the bottom.

Similarly, along the vertical side we'd have the h go to h+3+3 = h+6

The old rectangle that was h by h+4 is now h+6 by h+10

Multiply these expressions to find the area

area = length*width

area = (h+6)(h+10)

area = x(h+10) ..... replace h+6 with x

area = xh + 10x .... distribute

area = h( x ) + 10( x )

area = h( h+6 ) + 10( h+6 ) .... plug in x = h+6

area = h^2+6h + 10h+60 .... distribute again twice more

area = h^2 + 16h + 60

You can also use the box method or the FOIL rule as alternative routes to find the area.

The units for the area are in square inches.

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3 years ago