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solniwko [45]
3 years ago
11

a marble is rolled down a ramp. The distance it travels is described by the formula d=490T^2 where d is the distance in centimet

ers that the marble rolls in t seconds. If the marble is released at the top of a ramp that is 3,920 cm long for what time period will the marble be more than halfway down the ramp?
Mathematics
1 answer:
Helga [31]3 years ago
5 0

Answer:

The period when its more than halfway

= 2 < t  ≤ 2.83 seconds   ( to nearest hundredth).

Step-by-step explanation:

When is is halfway down the ramp d = 3920/2 = 1960 cm, so:

1960 = 490 t^2

t^2 = 1960/490 = 4

t = 2 seconds.

When it reaches the bottom d = 3920 :

t^2 = 3920/490 =  8

t = 2.83 seconds.

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kkurt [141]

Answer:

(x- \frac{2}{5})^{2} = x^2 -\frac{4}{5}x + \frac{4}{25}

Step-by-step explanation:

You have two methods to expand this binomial.

Method 1  

If you have the expression:

(x- \frac{2}{5})^{2}

You can write the expression it in the following way:

(x-\frac{2}{5})^{2}=(x-\frac{2}{5})(x-\frac{2}{5})

Then, apply the distributive property:

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Simplify the expression:

(x-\frac{2}{5})^2= x^2 -\frac{4}{5}x+ (\frac{4}{25})

---------------------------------------------------------------------------------------

Method 2

For any expression of the form:

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If

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(x- \frac{2}{5})^{2} = x^2 - 2x\frac{2}{5} + (\frac{2}{5})^2

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