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Gekata [30.6K]
3 years ago
11

What is the decimal equivalent of 9/12

Mathematics
2 answers:
dezoksy [38]3 years ago
8 0

\dfrac{9}{12}=\dfrac{9:3}{12:3}=\dfrac{3}{4}=\dfrac{3\cdot25}{4\cdot25}=\dfrac{75}{100}=0,75

tino4ka555 [31]3 years ago
5 0

Answer:

0.75

Step-by-step explanation:

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Simplify (-1 1/2) ÷ 3/4 x -2/3
Goryan [66]

Answer:

Step-by-step explanation:

1\frac{1}{2}=1×\frac{2}{2}, +\frac{1}{2}

(-\frac{3}{2})÷\frac{3}{4}×(-\frac{2}{3})

\frac{3}{2}×\frac{4}{3}×\frac{2}{3}

\frac{1}{2}×4×\frac{1}{3}

4×\frac{1}{3}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}

=\frac{4}{3}

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2 years ago
The function a(t)=t^(1/2)−t^(−1/2) m/s^2 represents the acceleration of a particle moving along a horizontal axis. At time t=0,
garri49 [273]

Answer:

see below

Step-by-step explanation:

a(t)=t^(1/2)−t^(−1/2)

We integrate to find the velocity

v(t) = integral t^(1/2)−t^(−1/2) dt

     = t ^ (1/2 +1)         t ^ (-1/2 +1)

          ------------   -    -----------------  + c  where c is the constant of integration

              3/2                   1/2

v(t) = 2/3 t^ 3/2  - 2 t^ 1/2 +c

We find c by letting t=0 since we know the velocity is 4/3 when t=0

v(0) = 2/3 0^ 3/2  - 2 0^ 1/2 +c = 4/3

       0+c =4/3

       c = 4/3

v(t) = 2/3 t^ 3/2  - 2 t^ 1/2 +4/3

To find the position function we need to integrate the velocity

p(t) = integral 2/3 t^ 3/2  - 2 t^ 1/2 +4/3 dt

     2/3 t ^ (3/2 +1)        2 t ^ (1/2 +1)           4/3t

          ------------   -    -----------------  + ------------- + c  

              5/2                   3/2                    1

p(t) =  4/15 t^ 5/2 - 4/3t ^ 3/2 + 4/3t +c

We find c by letting t=0 since we know the position is -4/15 when t=0

p(0) =  4/15 0^ 5/2 - 4/3 0 ^ 3/2 + 4/3*0 +c = -4/15

         0 +c = -4/15

            c = -4/15

p(t) =  4/15 t^ 5/2 - 4/3t ^ 3/2 + 4/3t -4/15

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Answer:

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Step-by-step explanation:

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