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Ivahew [28]
3 years ago
6

Find the value of x (LAST QUESTION)

Mathematics
2 answers:
astraxan [27]3 years ago
6 0

Answer:

9. x= 9

10. x= 9

Step-by-step explanation:

9. 13x+12= 129

    13x=129-12

       x=117/13=9

10.180-(7x-10)= 12x+19

     180-7x+10=12x+19

     190-7x=12x+19

     190=19x+19

     190= 19(x+1)

     190/ 19=x+1

     10=x+1

     9=x

nika2105 [10]3 years ago
4 0

Answer:

9. 13x+12 = 129

=> x=117/13 = 9

10.7x-10+(12x+19)= 180

19x+9= 180

x= 171/19

= 9

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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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