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3 years ago

Find the quotient of '(16t^2-4)/(8t+4)'

Mathematics

1 answer:

Marizza181 [45]3 years ago

3 0

here we have to find the quotient of '(16t^2-4)/(8t+4)'

now we can write 16t^2 - 4 as (4t)^2 - (2)^2

the above expression is equal to (4t + 2)(4t - 2)

there is another expression (8t + 4)

the expression can also be written as 2(4t + 2)

now we have to divide both the expressions

by dividing both the expressions we would get (4t + 2)(4t - 2)/2(4t + 2)

therefore the quotient is (4t - 2)/2

the expression comes out to be (2t - 1)

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user100 [1]

Answer:

It's position at time t = 5 is 593.

Step-by-step explanation:

The velocity v(t) is the integral of the acceleration a(t)

The position s(t) is the integral of the velocity v(t)

We have that:

The acceleration is:

$a(t) = 24t + 2$

Velocity:

$v(t) = \int {a(t)} \, dt = \int {24t + 2} \, dt = 12t^{2} + 2t + K$

K is the initial velocity, that is v(0). Since V(0) = 13, K = 13

Then

$v(t) = 12t^{2} + 2t + 13$

Position:

$s(t) = \int {s(t)} \, dt = \int {12t^{2} + 2t + 13} \, dt = 4t^{3} + t^{2} + 13t + K$

Since s(0) = 3

$s(t) = 4t^{3} + t^{2} + 13t + 3$

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This is s(5).

$s(t) = 4t^{3} + t^{2} + 13t + 3$

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$s(5) = 593$

It's position at time t = 5 is 593.

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