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devlian [24]
3 years ago
10

Find the accelartion:

Mathematics
1 answer:
bagirrra123 [75]3 years ago
6 0
A. Man runs (> 2.2 m/s is running) uphill at 6 m/s (???) and decelerates to walking at 2.0 m/s near the top, all in 5 minutes, assuming he ran in a straight line.

Acceleration is (final velocity-initial velocity (m/s) )/time (s)
=(2-6)/(5*60)=-4/300= -0.0133 m/s^2

B. Car goes east for 300 km in 45 min.
Velocity 
=300*1000 m  / (45*60) s  due east
=300000/2700 m/s due east
=111.1 m/s due east
(is it an airplane?  A cesna cruises at 63 m/s  (226 km/h) )
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Solve the following system of equations. What is one x-value of a solution?
gulaghasi [49]

Answer:

There is some mistake in the question, because the solutions are x = -1.445 and x = -34.555

Step-by-step explanation:

Given the functions:

f(x) = x² + 4x + 10

g(x) = -32x - 40

we want to find the points at which f(x) = g(x).

x² + 4x + 10 = -32x - 40

x² + 4x + 10 + 32x + 40 = 0

x² + 36x + 50 = 0

Using quadratic formula:

x = \frac{-b \pm \sqrt{b^2-4(a)(c)}}{2(a)}

x = \frac{-36 \pm \sqrt{36^2-4(1)(50)}}{2(1)}

x = \frac{-36 \pm 33.11}{2}

x_1 = \frac{-36 + 33.11}{2}

x_1 = -1.445

x_2 = \frac{-36 - 33.11}{2}

x_2 = -34.555

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(11 * 25) + (7.5 * 19) = 

$417.50 
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Answer: 43

Step-by-step explanation:

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CAN SOMEONE HELP .. WITH THE RIGHT ASWER PLZZ...
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