Y1 is the simplest parabola. Its vertex is at (0,0) and it passes thru (2,4). This is enough info to conclude that y1 = x^2.
y4, the lower red graph, is a bit more of a challenge. We can easily identify its vertex, which is (-4,0), and several points on the grah, such as (2,-3).
Let's try this: assume that the general equation for a parabola is
y-k = a(x-h)^2, where (h,k) is the vertex. Subst. the known values,
-3-(-4) = a(2-0)^2. Then 1 = a(2)^2, or 1 = 4a, or a = 1/4.
The equation of parabola y4 is y+4 = (1/4)x^2
Or you could elim. the fraction and write the eqn as 4y+16=x^2, or
4y = x^2-16, or y = (1/4)x - 4. Take your pick! Hope this helps you find "a" for the other parabolas.
Two vertical angles are formed. These angles are also equal to each other.
Yes you can! Say you have 5(2 + x) + 3x. The answer would be 10+8x because when you use the distributive property you will end up getting 5*2 5*x and once you multiply, it's obvious 5 times 2 is 10 and 5 times x will equal 5x. Then you add 5x and 3x and get 8x. So, the simplified expression here is 10+8x. When you have this as an answer it's proving your answer right. You can without using the distributive property. We can have 10+5x+3x and it will be your simplified expression. It's the same thing with equations too.
There were 16 states added.
Tennessee, Ohio, Louisiana, Indiana, Mississippi, Illinois, Alabama, Maine,
Missouri, Arkansas, Michigan, Florida, Texas, Wisconsin, and California
