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.
Set up
Let the dimes = d
Let the pennies = p
Let the quarters = q
Equations
You cannot mean that the pennies and dimes have equal numbers. That would mean that each had 21.5 members. Now could you mean that the dime and penny amount could be the same with 43 coins that total 4.00. Four dollars means that you need 40 dimes alone. It must mean that you are including quarters.
p + d + p = 43 (1)
p = d (2)
p +10d +25q = 451 (3)
Note how this last equation = was derived. You have to multiply the dimes by 10 and he quarters by 100 and the total by 100 to get the numbers all in pennies.
Put the results of 2 into 1.
2p + q = 43 (4)
You need to modify equation 3 as well.
p + 10p + 25q = 451
11p + 25q = 451 (5)
Solve the new equations
2p + q = 43 (4)
11p + 25q = 451 (5)
Multiply 4 by 25
25(2p- + q = 43)
50p + 25q = 1075 (6) Subtract (5) from (6)
<u>11p + 25q = 451
</u>39p = 624 Divide by 39
p = 624 / 39
p = 16
Since the pennies and dimes are equal there are 16 dimes
p + d + q = 43
16 + 16 + q = 43
32 + q = 43
q = 11
Check
16 + 10*16 + 11*25 = ?
16 + 160 + 275 = ?
451 = ?
Nice problem. Thanks for posting.
It goes on forever basically, as it can't be expressed as a fraction.
1 to 4
150mm is the same as 15cm
15cm:60cm
both divisible by 15
1cm:4cm
1 to 4
80 1.5 because 10 times 8 equals 80 plus 1.5 equals 80.1.5
