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spayn [35]
3 years ago
6

A length of string that is 22 feet long is being cut into pieces that are 1/3 foot long. how many pieces will there be

Mathematics
1 answer:
Anika [276]3 years ago
4 0
One foot is equivalent to 12 inches. If we get 1/3 of 12 inches, then it would be 4 inches. Given that the length of the string is 22 feet long, we first have to multiply 22 by 12 and we get 264 inches. Then divide this one by 4, so the number of pieces there will be is 66. Hope this answers your question.
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ohaa [14]

Answer:

0.15

Step-by-step explanation:

i goog.led it its probably right

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Please answer all part correct​
insens350 [35]

Answer:

a). We want to know how much each point was worth.

b). 12x+5=41

c). Each problem worth 3 points.

Step-by-step explanation:

a). We want to know how much each problem was worth. Because we have the total points of the test, and how much was the bonus. But we still don't know the worth of each problem.

b). We know that the total points of the test were 41, and the bonus 5 points. There were 12 problems on the test and we are going to use "x" for the unknown part (how many points each problem was worth).

The equation is :

12x+5=41

Why 12x? Because if you multiply the twelve problems of the test with the worth of each one and then add the 5 points of the bonus you will obtain the total points of the test (41).

c). Now we have to solve the equation, this means that we have to clear "x":

12x+5=41

Subtract 5 from both sides.

12x+5-5=41-5\\12x=36

Finally divide in 12 both sides of the equation:

\frac{12x}{12}=\frac{36}{12} \\\\x=3

Then each problem worth 3 points.

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Express in scientific notation 472.3 in decimal form
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If you minus a positive number from a negative number is the diffirance a positive number, or a negative number? For example: 5
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Step-by-step explanation:

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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

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