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geniusboy [140]
3 years ago
13

HELP ASAP PLEASE

Mathematics
1 answer:
dimaraw [331]3 years ago
7 0

Answer:

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate

10%. The second car depreciates at an annual rate of 15%. What is the approximate difference in the ages of the two

cars?

Step-by-step explanation:

the answer is in the question

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A ball is thrown from an initial height of 7 feet with an initial upward velocity of 24 fts. The ball's height h (in feet) after
Montano1993 [528]

Answer:

The ball's height is 11 feet at t = 0.19s and t = 1.31s

Step-by-step explanation:

The ball's height h (in feet) after seconds is given by the following function:

h(t) = 7 + 24t - 16t^{2}

Find all values of t for which the ball's height is 11 feet.

This is t when h(t) = 11.

We will need to solve a quadratic equation for this.

Solving a quadratic equation

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = (x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this problem:

h(t) = 11

7 + 24t - 16t^{2} = 11

16t^{2} - 24t + 4 = 0

Simplifying by 4

4t^{2} - 6t + 1 = 0

So a = 4, b = -6, c = 1

\bigtriangleup = (-6)^{2} - 4*(4)*1 = 20

x_{1} = \frac{-(-6) + \sqrt{20}}{2*4} = 1.31

x_{2} = \frac{-(-6) - \sqrt{20}}{2*4} = 0.19

The ball's height is 11 feet at t = 0.19s and t = 1.31s

7 0
3 years ago
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