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3 years ago

12

Speedie ran the 200 meter race in 25.5 seconds. How much of an hour did it take Speedie to run

1 answer:

grigory [225]3 years ago

5 0

Answer:

Speedie ran the race in 0.708(3)% of an hour, the 3 in the parentheses means the 3 repeats forever

Step-by-step explanation:

We have 25.5/3600, and when we convert it to a decimal, we have 0.00708(3), the 3 in the parentheses means the 3 repeats forever. To convert it to percentage, we multiply it by 100, to get 0.708(3). Hope it helps!

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Bas_tet [7]

9514 1404 393

Answer:

x° = 29°

Step-by-step explanation:

The mnemonic SOH CAH TOA is intended to remind you of the relevant trig relationships. Here, you have the hypotenuse and the side opposite the unknown angle. So, the appropriate relation is ...

Sin = Opposite/Hypotenuse

sin(x°) = 12/25

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The value of x is about 29.

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2 years ago

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3 years ago

In physics, if a moving object has a starting position at so, an initial velocity of vo, and a constant acceleration a, the

emmasim [6.3K]

Answer:

$a=\frac{2S -2v_ot-2s_o}{t^2}$

Step-by-step explanation:

We have the equation of the position of the object

$S = \frac{1}{2}at ^2 + v_ot+s_o$

We need to solve the equation for the variable a

$S = \frac{1}{2}at ^2 + v_ot+s_o$

Subtract $s_0$ and $v_0t$ on both sides of the equality

$S -v_ot-s_o = \frac{1}{2}at ^2 + v_ot+s_o - v_ot- s_o$

$S -v_ot-s_o = \frac{1}{2}at ^2$

multiply by 2 on both sides of equality

$2S -2v_ot-2s_o = 2*\frac{1}{2}at ^2$

$2S -2v_ot-2s_o =at ^2$

Divide between $t ^ 2$ on both sides of the equation

$\frac{2S -2v_ot-2s_o}{t^2} =a\frac{t^2}{t^2}$

Finally

$a=\frac{2S -2v_ot-2s_o}{t^2}$

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3 years ago

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jeyben [28]

Answer:False

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3 years ago