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Novosadov [1.4K]

3 years ago

8

I have a 18x32 inch Sheet of paper, I would like to know how to put 30 squares evenly on the paper. How many inches tall and how

many wide would each square be

1 answer:

GaryK [48]3 years ago

3 0

Answer:

2plus 2 is 4

Step-by-step explanation:

4 minus 1 thats 3 quick maths

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Find the product:<br><br> 5(−3)(−2)

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3 years ago

WILL MARK BRAINLIEST!!! 40 POINTS!! ACTUAL ANSWERS, PLZZZ

steposvetlana [31]

Answer:

Part A:

$\left(x + 7\right)^{5}=x^{5} + 35 x^{4} + 490 x^{3} + 3430 x^{2} + 12005 x + 16807$

Part B:

The closure property describes cases when mathematical operations are CLOSED. It means that if you apply certain mathematical operations in a polynomial it will still be a polynomial. Polynomials are closed for sum, subtraction, and multiplication.

It means:

$\text{Sum of polynomials } \Rightarrow \text{ It will always be a polynomial}$

$\text{Subtraction of polynomials } \Rightarrow \text{ It will always be a polynomial}$

$\text{Multiplication of polynomials } \Rightarrow \text{ It will always be a polynomial}$

But when it is about division:

$\text{Division of polynomials } \Rightarrow \text{ It will not always/sometimes be a polynomial}$

<u>Example of subtraction of polynomials:</u>

<u /> $(2x^2+2x+3) - (x^2+5x+2)$ <u />

<u /> $x^2-3x+1$ <u />

Step-by-step explanation:

<u>First, it is very important to define what is a polynomial in standard form: </u>

It is when the terms are ordered from the highest degree to the lowest degree.

Therefore I can give:

$x^5-5x^4+3x^3-3x^2+7x+20$

but,

$x^5+3x^3-3x^2+7x+20-5x^4$ is not in standard form.

For this question, I can simply give the answer: $x^5-5x^4+3x^3-3x^2+7x+20$ and it is correct.

But I will create a fifth-degree polynomial using this formula

$(a+b)^n=\sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$

Also, note that

$\binom{n}{k}=\frac{n!}{(n-k)!k!}$

For $a=x \text{ and } b=7$

$\left(x + 7\right)^{5}=\sum_{k=0}^{5} \binom{5}{k} \left(x\right)^{5-k} \left(7\right)^k$

$\text{Solving for } k \text{ values: } 0, 1, 2, 3, 4 \text{ and } 5$

Sorry but I will not type every step for each value of $k$

The first one is enough.

For $k=0$

$\binom{5}{0} \left(x\right)^{5-0} \left(7\right)^{0}=\frac{5!}{(5-0)! 0!}\left(x\right)^{5} \left(7\right)^{0}=\frac{5!}{5!} \cdot x^5= x^{5}$

Doing that for $k$ values:

$\left(x + 7\right)^{5}=x^{5} + 35 x^{4} + 490 x^{3} + 3430 x^{2} + 12005 x + 16807$

8 0

3 years ago