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Juliette [100K]
3 years ago
6

I need help please anyone?????!!!

Mathematics
2 answers:
melomori [17]3 years ago
8 0

▪▪▪▪▪▪▪▪▪▪▪▪▪  {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪

Here we go ~

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:6(x - 2)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:(6 \times x) - (6 \times 2)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:(6x) - (12)

{ \qquad{ \sf{ \dashrightarrow}}}  \:  \: \sf \:6x - 12

Therefore the Correct choice is ~

\overbrace{  \underbrace{\underline{ \boxed{ \sf c}}}}

Andreyy893 years ago
5 0

Answer:

6x-12

Step-by-step explanation:

6(x-2)

6x -12 ...just open bracket and multiply the following that was inside the bracket

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Mr Thompson please help with my new HW
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Answers:

  1. C) x = plus/minus 11
  2. B) No real solutions
  3. C) Two solutions
  4. A) One solution
  5. The value <u>  18  </u> goes in the first blank. The value <u>  17  </u> goes in the second blank.

========================================================

Explanations:

  1. Note how (11)^2 = (11)*(11) = 121 and also (-11)^2 = (-11)*(-11) = 121. The two negatives multiply to a positive. So that's why the solution is x = plus/minus 11. The plus minus breaks down into the two equations x = 11 or x = -11.
  2. There are no real solutions here because the left hand side can never be negative, no matter what real number you pick for x. As mentioned in problem 1, squaring -11 leads to a positive number 121. The same idea applies here as well.
  3. The two solutions are x = 0 and x = -2. We set each factor equal to zero through the zero product property. Then we solve each equation for x. The x+2 = 0 leads to x = -2.
  4. We use the zero product property here as well. We have a repeated factor, so we're only solving one equation and that is x-3 = 0 which leads to x = 3. The only root is x = 3.
  5. Apply the FOIL rule on (x+1)(x+17) to end up with x^2+17x+1x+17 which simplifies fully to x^2+18x+17. The middle x coefficient is 18, while the constant term is 17.
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