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Otrada [13]
3 years ago
9

(Adding and Subtracting Polynomials)

Mathematics
1 answer:
andrey2020 [161]3 years ago
6 0

Adding and subtracting big polynomials like these are pretty easy. You just need to combine like terms. For example:

1.)

{5x}^{2}  +  {3x}^{2}  =  {8x}^{2}

2.)

({3x}^{2}  + 5xy) + (7xy + 2) =

{3x}^{2} + 12xy + 2

(The 3x^2 and the 2 stay intact while the 5xy and 7xy combine together)

All you have to do is combine the numbers that have the same powers of x and y with each other. x^2 will combine with x^2 and xy^2 wil combine with xy^2 exc. If there is no other number with the same x and y's, then you just leave it as it is in the answer.

Now with the original question, I see a -9xy^3, and thats gonna combine with the 3xy^3 in the second polynomial and the 2xy^3 in the third one.

- 9x {y}^{3}  + 3x {y}^{3}  + 2x {y}^{3} = \\  - 4x {y}^{3}

So far we have -4xy^3, the next term is going to be a -9x^4y^3, and that's gonna combine with the 3x^4y^3 in the third one.

- 9 {x}^{4} {y}^{3}  + 3 {x}^{4}  {y}^{3}  =  - 6{x}^{4}  {y}^{3}

We now finished adding the like terms that were in the first polynomial, we will move onto the second polynomial. The first term in this one is 3xy^3, in which we already added in the first step. At this point, it doesn't look like there are any other terms that have the same x and y behind them. So we can move on and write the final answer:

- 4x {y}^{3} - 6 {x}^{4} {y}^{3}  + 7 {y}^{4}  \\ - 8 {x}^{4} {y}^{4}

(All on the same line of course)

Also, for your second question, the order does not matter in which you write the terms. I could write the 7y^4 behind the -8x^4y^4 and it would still be the same answer.

If you have any other questions let me know :) while I double check my work.

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Answer:

Part A

f(n)=52-12(n-1)

f(n)=\left\{\begin{matrix}52\: \:if \: \:n=1 & \\f(n+1)+12& if\: n\geq 2 \end{matrix}\right.

Part B

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Part C

1/4,3/4,5/4,7/4,9/4

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Part D:

h(n)=1.1+0.4(n-1)\\h(n)=\left\{\begin{matrix}1.1 & if\:n=1 \\ h(n+1)+0.4 & if\:n\geq 2\end{matrix}\right

Step-by-step explanation:

By definition, an Arithmetic Sequence holds the same difference between each following number.

Part A

(52,40, 28, 16)\\52-40=12\\40-28=12\\28-16=12\\d=12

<u>Explicit Formula</u>

To write an explicit formula is to write it as function.

f(n)=52-12(n-1)

<u>Recursive Formula</u>

To write it as recursive formula, is to write it as recurrence given to some restrictions:

f(n)=\left\{\begin{matrix}52\: \:if \: \:n=1 & \\f(n+1)+12& if\: n\geq 2 \end{matrix}\right.

Part B

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Geometric Sequence, since 2*2=4 8*2=16 and 16*2=32 and 8+2=10 8+16=24

Part C

(\frac{1}{4},\frac{3}{4},\frac{5}{4},\frac{7}{4},\frac{9}{4})\\\

Arithmetic Sequence, difference

d=\frac{2}{4}

<u>Explicit Formula:</u>

g(n)=\frac{1}{4}+\frac{2}{4}(n-1)

<u>Recursive Formula</u>

g(n)=\left\{\begin{matrix}\frac{1}{4} &if\:n=1 \\ g(n+1)+\frac{2}{4} &if\: n\geq 2\end{matrix}\right.

Part D

(1.1,1.5,1.9,2.3,2.7) Arithmetic Sequence, difference d=0.4

<u>Explicit formula</u>

h(n)=1.1+0.4(n-1)\\

<u>Recursive Formula</u>

h(n)=\left\{\begin{matrix}1.1 &if\:n=1 \\ h(n+1)+0.4 &if\: n\geq 2\end{matrix}\right.

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