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yulyashka [42]
3 years ago
12

SOMEONE HELP ME PLEASE ILL GIVE BRAINLY

Mathematics
1 answer:
e-lub [12.9K]3 years ago
8 0

Answer:

Step-by-step explanation:

In order to determine the information you're being asked for, you need to complete the square on that quadratic. The first step is to move the constant over to the other side of the equals sign:

x^2-2x=3

Here would be the step where, if the leading coefficient isn't a 1, you'd factor it out. But ours is a 1, so we're good there. Now take half the linear term (the term with the single x on it), square it, and add it to both sides. Our linear term is a -2.  Half of -2 is -1, and -1 squared is +1. We add +1 to both sides giving us this:

x^2-2x+1=3+1

Now we'll clean it up a bit. The right side becomes a 4, and the left side is written as its perfect square binomial, which is the whole reason we did this. That binomial is

(x-1)^2=4 (set equal to the 4 here). Now we'll move the 4 back over and set the whole thing back equal to y:

y=(x-1)^2-4

From this it's apparent what the vertex is: (1, -4),

the axis of symmetry is x = 1, and

the y-intercept is found by setting the x's equal to 0 in the original  equation and solving for y. So the y-intercept is (0, -3).

Your choice for the correct answer is the very last one there.

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<h2>(-10, 2, 6)</h2>

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\left\{\begin{array}{ccc}x+3y+2z=8&(1)\\3x+y+3z=-10&(2)\\-2x-2y-z=10&(3)\end{array}\right\qquad\text{subtract both sides of the equations (1) from (2)}\\\\\underline{-\left\{\begin{array}{ccc}3x+y+3z=-10\\x+3y+2z=8\end{array}\right }\\.\qquad2x-2y+z=-18\qquad(4)\qquad\text{add both sides of the equations (3) and (4)}\\\\\underline{+\left\{\begin{array}{ccc}-2x-2y-z=10\\2x-2y+z=-18\end{array}\right}\\.\qquad-4y=-8\qquad\text{divide both sides by (-4)}\\.\qquad\qquad y=2\qquad\text{put the value of y to (1) and (3)}

\left\{\begin{array}{ccc}x+3(2)+2z=8\\-2x-2(2)-z=10\end{array}\right\\\left\{\begin{array}{ccc}x+6+2z=8&\text{subtract 6 from both sides}\\-2x-4-z=10&\text{add 4 to both sides}\end{array}\right\\\left\{\begin{array}{ccc}x+2z=2&\text{multiply both sides by 2}\\-2x-z=14\end{array}\right\\\underline{+\left\{\begin{array}{ccc}2x+4z=4\\-2x-z=14\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad\qquad3z=18\qquad\text{divide both sides by 3}\\.\qquad\qquad z=6\qquad\text{put the value of z to the first equation}

x+2(6)=2\\x+12=2\qquad\text{subtract 10 from both sides}\\x=-10

4 0
3 years ago
The quotient of a number and - 2/3 is -9/10.
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It would be: -2/3 * -9/10 = 18/30 = 3/5

So, option B is your answer.

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