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astra-53 [7]
3 years ago
5

Consider the problem: After charging for 1/3 of an hour, a phone is at 2/5 of its full power. How long will it take the phone to

charge completely? Decide whether each equation can represent the situation.

Mathematics
2 answers:
prisoha [69]3 years ago
5 0

Answer:

Equation 2

is the answer

Step-by-step explanation:

Alinara [238K]3 years ago
4 0

To calculate the time for a full charge, you would divide the time already charged by the amount of charge:

1/3 / 2/5 = 1/3 x 5/2 = 5/6 of an hour for full charge.

You could use 2 and 4 to find the time.

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The distance of a golf ball from the hole can be represented by the right side of a parabola with vertex (-1,8). The ball reache
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Answer:

Let's suppose that the hole is at y = 0m, where x is the time variable.

we know that:

The vertex is (-1s, 8m).

(i suppose x in seconds and y in meters)

At x = 1s, the ball reaches the hole, so we also have the point:

(1s, 0m).

Remember that the vertex of a quadratic equation y = a*x^2 + b*x + c is at:

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then we have:

-1 = -b/2a.

Then we have tree equations:

8m = a*(-1s)^2 + b*-1s + c

0m = a*(1s)^2 + b*1s + c

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First we should isolate one variable in the third equation, and then replace it in one of the other two:

1s*2a = b.

So we can replace b in the first two equations bi 1s*2a.

8m = a*1s^2 - 1s*2a*1s + c

0m = a*1s^2 + 1s*2a*1s + c

We can simplify both equations and get:

8m = a*( 1s^2 - 2s^2) + c = -a*1s^2 + c.

0m = a*(1s^2 + 2s^2) + c = a*3s^2 + c.

Easily we can isolate c in the second equation and then replace it into the first equation:

c = -a*3s^2

The first equation becomes:

8m = -a*1s^2 - a*3s^2 = -a*4s^2

a = 8m/-4s^2 = -2m/s^2.

Now with a, we can find the values of c and b.

c = -a*3s^2 = -(-2m/s^2)*3s^2 = 6m.

b = 1s*2a = 1s*(-2m/s^2) = -2m/s.

Then the equation is:

y = (-2m/s^2)*t^2 + (-2m/s)*t + 6m

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