Can you help me with this question. The relationship between d and t is of the form d=at+bt^2 where a and b?

The relationship between d and t is of the same form
d=at+bt^2
where a and b are constants
use the ordered pairs (2,11) and (4,28) to calculate the values of a and b.

Answer:
Try this:

When the point is (2,11),

2a + 4b = 11

When the point is: (4, 28),

4a + 16b = 28

Now multiply the entire first equation by 2 to get:

4a + 8b = 22

Now deduct the second equation from this one to get:
8b = 6 ====> b = 3/4

When b = 3/4 in the very first equation you get:

2a + (4 x 3/4) = 11
===> 2a = 8
===> a = 4

Therefore, a = 4 and b = 3/4 is the answer.

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(t, d) = (2,11), (4, 28)

11 = a(2) + b(2^2) = 2a +4b
28 = a(4) + b(4^4) = 4a +16b

now, solve the system of equations:
from first equation,
22 = 4a +8b
4a = 22 - 8b

now, substitute into second equation
28 = 22 - 8b +16b
6 = 8b
b = 3/4

substitute this value back into either equation:
11 = 2a +4(3/4)
11 = 2a + 3
2a = 8
a = 4

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Can you help me with this question. The relationship between d and t is of the form d=at+bt^2 where a and b?

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