All categories

2 years ago

John’s solution to an equation is shown below.

Mathematics

1 answer:

g100num [7]2 years ago

8 0

C would be the correct option

You might be interested in

1. 8y^2-4y+1 divided by 2y-1

____ [38]

Q1. The answer is $4y+ \frac{1}{2y-1}$

$\frac{8y^{2}-4y+1 }{2y-1} = \frac{4y*2y-4y+1}{2y-1} = \frac{4y(2y-1)+1}{2y-1} = \frac{4y(2y-1)}{2y-1}+ \frac{1}{2y-1} =4y+ \frac{1}{2y-1}$

Q2. The answer is $2a+3+ \frac{6}{a-1}$

$\frac{2 a^{2}+a+3 }{a-1} = \frac{a*2a-2a+3a+3}{a-1} = \frac{2a(a-1)+3a+3}{a-1}= \frac{2a(a-1)}{a-1}+ \frac{3a+3}{a-1} = \\ \\ =2a+ \frac{3a+3}{a-1}=2a+ \frac{3a-3+3+3}{a-1}=2a+ \frac{3(a-1)+6}{a-1} =2a+ \frac{3(a-1)}{a-1} + \frac{6}{a-1}= \\ \\ =2a+3+ \frac{6}{a-1}$


Q3. The answer is $2 x^{2} +5x+2$ 

 $\frac{6 x^{3} +11 x^{2} -4x-4}{3x-2} = \frac{3x*2 x^{2}-2 x^{2} *2+15 x^{2} -4x-4 }{3x-2} = \frac{2 x^{2} (3x-2)+15 x^{2} -4x-4}{3x-2}= \\ \\ = \frac{2 x^{2} (3x-2)}{3x-2} + \frac{15 x^{2} -4x-4}{3x-2} =2 x^{2} +\frac{15 x^{2} -4x-4}{3x-2}=2 x^{2} + \frac{15 x^{2} -10x+6x-4}{3x-2}= \\ \\ =2 x^{2} + \frac{5x*3x-5x*2+6x-4}{3x-2} =2 x^{2} + \frac{5x(3x-2)+3x*2-2*2}{3x-2} = \\ \\ =2 x^{2} + \frac{5x(3x-2)}{3x-2} + \frac{3x*2-2*2}{3x-2} =2 x^{2} +5x+ \frac{2(3x-2)}{3x-2} =2 x^{2} +5x+2$


Q4. The answer is 2x + 7

 $\frac{6 x^{2} +11x-35}{3x-5} = \frac{6 x^{2} -10x+21x-35 }{3x-5} = \frac{3 x *2x-5*2x+7*3x-7*5 }{3x-5} = \\ \\ = \frac{2x(3x-5)+7(3x-5)}{3x-5}= = \frac{(3x-5)(2x+7)}{3x-5} =2x+7$


Q5. The answer is $x+1- \frac{3}{x-1}$ 

 $\frac{ x^{2} -4}{x-1} = \frac{ x^{2} -x+x-1-3 }{x-1} = \frac{x*x-x+x-1-3}{x-1} = \frac{x(x-1)+(x-1)-3}{x-1} = \\ \\ \frac{(x+1)(x-1)-3}{x-1} = \frac{(x+1)(x-1)}{x-1} -\frac{3}{x-1} =x+1- \frac{3}{x-1}$

Q6. The answer is $y^{2} -2y+3$

$\frac{ y^{3}-4 y^{2}+7y-6 }{y-2} = \frac{y* y^{2} -2y^{2}-2 y^{2} +7y-6 }{y-2} = \frac{y^{2}(y-2)-2 y^{2} +7y-6}{y-2}= \\ \\ = \frac{y^{2}(y-2)}{y-2}+ \frac{-2 y^{2} +7y-6}{y-2} = y^{2} + \frac{-2 y^{2} +4y + 3y-6}{y-2} = \\ \\ =y^{2} + \frac{-2y*y-2y(-2)+3y-3*2}{y-2} = y^{2} + \frac{(-2y)(y-2)+3(y-2)}{y-2} = \\ \\ = y^{2} + \frac{(-2y+3)(y-2)}{y-2} = y^{2} +(-2y+3) =y^{2} -2y+3$


Q7. The answer is $x^{2} +xy+ y^{2}}{x-y}$ 

 $\frac{ x^{3} - \frac{x}{y} y^{3} }{x-y} = \frac{(x-y)( x^{2} +xy+ y^{2}) }{x-y} = \frac{ x^{2} +xy+ y^{2}}{x-y}$


Q8. The answer is $(a^{2} +2ab+2b^{2})$ 

 $\frac{a^{4} +4b^{4} }{a^{2}-2ab+2 b^{2} } = \frac{ (a^{2})^{2} +(2b)^{2}}{a^{2}-2ab+2 b^{2}} = \frac{(a^{2} -2ab+2b^{2})(a^{2} +2ab+2b^{2}) }{(a^{2} -2ab+2b^{2})} =(a^{2} +2ab+2b^{2})$ 

Q9. The answer is a^{n-8} - a^{-14}

 $\frac{ (a^{2}) ^{n} - a^{n-6} }{ a^{n+8} }= \frac{(a^{2}) ^{n} }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}} \\ \\ (x^{y}) ^{z}= x^{y*z} \\ \\ \frac{ x^{y} }{ x^{z} } = x^{y-z} \\ \\ 
\frac{(a^{2}) ^{n} }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}} =\frac{a^{2n} }{a^{n+8}}- \frac{ a^{n-6} }{a^{n+8}} = a^{2n-(n+8)} - a^{n-6-(n+8)} = \\ \\ =a^{2n-n-8} - a^{n-6-n-8} = a^{n-8} - a^{-14}$

5 0

3 years ago

What is the value of x? Enter your answer in the box.

Shkiper50 [21]

Step-by-step explanation:

114° thats what i got when i calculated

8 0

2 years ago

Read 2 more answers

ryan earns $5 for each small dog he walks and $8 for each latge dog he walks. today he walked 8 dogs and made a total of $55. ho

ad-work [718]

If we divide 55 (the total amount of money he made) by 5 (price to walk one small dog), we get 11. Meaning that the maximum amount of small dogs he could possibly walk and receive that amount of money is 11. However, he only walked 8 dogs. Therefore Ryan would have needed to walk a specific amount of large dogs in order to earn 55 dollars in total.

After doing some process of elimination, I reached this conclusion:

3 small dogs = $15

5 large dogs = $40

The combination of the two would equal $55.

Therefore, the answer would be 3 small dogs.

7 0

3 years ago

Mr. Bernard needs to order more boxes for his 16 inch diameter “Super Pizza.” Use 3.14 to approximate π.

Olegator [25]

Approximately 1.49323

6 0

2 years ago

How do you find the x i need help really fast

kvasek [131]

I think that answer would be C

5 0

3 years ago