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2 years ago

Find the length of the variable .

Mathematics

1 answer:

ira [324]2 years ago

5 0

DONT CLICK THE LINK!!!

THEY WILL FIND YOU ADDRESS

I REPEAT DONT CLICK THE LINK!!!!!!

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need help with this someone please help

qwelly [4]

Answer: B

Step-by-step explanation:

7 0

3 years ago

What adds to make -4 but times to get -12

dimulka [17.4K]

3 because 4×3=12 and if you divide12÷3/4 it helps you check

6 0

3 years ago

Can't anybody help me with this ik math sucks but help

Agata [3.3K]

Answer:

Step-by-step explanation:

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2 years ago

What is thd solution for the equation 14.8=n-0.3

Nina [5.8K]

The given equation is:

14.8 = n - 0.3

Adding 0.3 to both sides, we get:

14.8 + 0.3 = n - 0.3 + 0.3
15.1 = n - 0
15.1 = n

Thus, the value of n for which the given equation is true is 15.1

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3 years ago

A ball moves in a straight line has an acceleration of a(t) = 2t + 5. Find the position function of the ball if its initial velo

Vlad1618 [11]

Answer:

$s(t) = frac{t^3}{3} + \frac{5t^2}{2} - 3t + 12$

Step-by-step explanation:

Relation between acceleration, velocity and position:

The velocity function is the integral of the acceleration function.

The position function is the integral of the velocity function.

Acceleration:

As given by the problem, the acceleration function is $a(t) = 2t + 5$

Velocity:

$v(t) = \int a(t) dt = \int (2t+5) dt = \frac{2t^2}{2} + 5t + K = t^2 + 5t + K$

In which K is the constant of integration, which is the initial velocity. So K = -3 and:

$v(t) = t^2 + 5t - 3$

Position:

$s(t) = \int v(t) dt = \int (t^2 + 5t - 3) = \frac{t^3}{3} + \frac{5t^2}{2} - 3t + K$

In which K, the constant of integration, is the initial position. Since it is 12:

$s(t) = frac{t^3}{3} + \frac{5t^2}{2} - 3t + 12$

4 0

3 years ago