Equation of a parabola with vertex at (2, -1) is
y = a(x - 2)^2 - 1
Using the given point: -3 = a(4 - 2)^2 - 1
-2 = a(2)^2
4a = -2
a = -1/2
Therefore, required equation is
y = -1/2(x - 2)^2 - 1
y = -1/2(x^2 - 4x + 4) - 1
y = -1/2x^2 + 2x - 2 - 1
y = -1/2x^2 + 2x - 3
For question 4 the Y needs to be by itself. So instead of it being x+2y>/-12 it would be y>/-6+x. You would have to subtract x on both sides and then divide 2 on both sides as well. whatever you do on one side, you have to do it on the other side. Hope this helps. After you redo all the equations, you graph them. If you need anymore help , let me know!
Answer:
(1, 2)
Step-by-step explanation:
Remember that the final shape and position of a figure after a transformation is called the image, and the original shape and position of the figure is the pre-image.
In our case, our figure is just a point. We know that after the transformation T : (x, y) → (x + 3, y + 1), our image has coordinates (4, 3).
The transformation rule T : (x, y) → (x + 3, y + 1) means that we add 3 to the x-coordinate and add 1 to the y-coordinate of our pre-image. Now to find the pre-image of our point, we just need to reverse those operations; in other words, we will subtract 3 from the x-coordinate and subtract 1 from the y-coordinate.
So, our rule to find the pre-image of the point (4, 3) is:
T : (x, y) → (x - 3, y - 1)
We know that the x-coordinate of our image is 4 and its y-coordinate is 3.
Replacing values:
(4 - 3, 3 - 1)
(1, 2)
We can conclude that our pre-image is the point (1, 2).
So this is going to be alot of writing to show my thinking but ill bold the answer.
1,1
1,2
1,3
1,4
1,5
2,1
2,2
2,3
2,4
2,5
3,1
3,2
3,3
3,4
3,5
4,1
4,2
4,3
4,4
4,5
5,1
5,2
5,3
5,4
5,5
next ill mark all the ones that equal 4 or 8 when added together, with an x
1,1
1,2
x1,3
1,4
1,5
2,1
x2,2
2,3
2,4
2,5
x3,1
3,2
3,3
3,4
x3,5
4,1
4,2
4,3
x4,4
4,5
5,1
5,2
x5,3
5,4
5,5
that is 6 (that equal 4 or 8) out of 25
so your ratio would be 6:19
It is more than just a quadrilateral. In fact it is going to be hard to pick.
These facts suit a square, a rectangle, a rhombus, and a parallelogram. And the above statement is true, but maybe a little harder to prove than the converse of the statement, which is the usual one you find.
The converse is "If you have a parallelagram, the diagonals bisect each other."
You might think a trapezoid deserves some mention. The diagonals of a trapezoid do not bisect each other.