# 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.

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.

Hope this helps and if so don't forget to vote for it!!

Now if anyone can answer this, they should get a medal, and well done to them.....

How are you ment to learn if you get others to do your homework for you!!

(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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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.

Hope this helps and if so don't forget to vote for it!!

Now if anyone can answer this, they should get a medal, and well done to them.....

How are you ment to learn if you get others to do your homework for you!!

(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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