What is the first step in solving this system by elimination?

1 answer:

4 0

You want to try and get rid of one of the variables ( x or y) so you would multiply the second equation by 2 which would make the x variable -2x which then when added to the first equation would eliminate the x variable.

Answer: c. Multiply the second equation by 2.

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Describe does Glide Reflection from the preimage triangle and black to the image triangle in red

AveGali [126]

Answer:

Step-by-step explanation:

Lets look at V which is at (2,-4).

We will also look at it's image V' which is at (-2,-1).

Let's go through the choices:

a. translation left 4 units and down 1 unit reflection x axis

Translating V left 4 and down 1 would give us (2-4,-4-1)

(-2,-5)

Reflecting that across the x-axis would give us

(-2,5).

So this is not (-2,-1) so let's move on...

b. translation left 2 units and up 2 units reflection line x=1

Translating V left 2 and up 2 gives us

(2-2,-4+2)

(0,-2)

Now reflecting this over x=1 gives us

(2,-2). ( Imagine this x=1 which is verical going right smack between points (0,-2) and (2,-2).)

We still did not reach our goal image.

c. translation left 2 units up and up 3 units reflection y axis

Starting at V(2,-4) and moving left 2 & up 3 gives (2-2,-4+3)=(0,-1).

Now reflecting this over the vertical axis, the y-axis, gives us (0,-1).

So we did not reach our (-2,-1) point yet.

d. translation left 2 units and 3 units reflection line x=-1.

Starting at (2,-4) then moving left 2 and up 3 gives us (2-2,-4+3)=(0,-1).

Now we are reflecting over x=-1 which means we want to choose a point on the side making x=-1 the middle. Thuis point on the other side of (0,-1) of x=-1 would be (-2,-1) which is the goal.

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3 years ago

I need help.. with this worksheet

ipn [44]

8. $298 4. 250 9. 28 6. 8.7 5. 465

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2 years ago

Find the zeroes of the following equation:

weqwewe [10]

Answer: $\bold{x=\dfrac{1\pm \sqrt5}{2}}$

both zeros are irrational numbers

<u>Step-by-step explanation:</u>

Note: The question would have made more sense if if it was 2/x = x - 1 but I will answer it as written.

$\dfrac{1}{x}=x-1\qquad \text{Restriction: }x\neq 0\\\\\\\text{Cross multiply, simplify, and set equal to zero:}\\1=x(x-1)\\\\1=x^2-x\\\\0=x^2-x-1\\\\\\\text{This is not factorable so you will need to use Quadratic Formula:}\\\\x=\dfrac{-(-1)\pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}\\\\\\x=\dfrac{1\pm \sqrt5}{2}$