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Naya [18.7K]
3 years ago
15

Mr. Brackett works in factory with his two sons. He is really happy with his job because he gets to spend some time with his bel

oved sons not only at home but also at work! He is allowed to take a break every 140 minutes while his two sons are allowed to take breaks in 210 minutes and 280 minutes. How many minutes will they have to wait after their first break together to get together again?
Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
3 0

Answer:

840 minutes

Step-by-step explanation:

We solve this question using the LCM Method

Find and list multiples of each number of minutes which is given in the quest as 140, 210 and 280 minutes until the first common multiple is found. This is the lowest common multiple.

Multiples of 140:

140, 280, 420, 560, 700, 840, 980, 1120

Multiples of 210:

210, 420, 630, 840, 1050, 1260

Multiples of 280:

280, 560, 840, 1120, 1400

Therefore,

LCM(140, 210, 280) = 840

Therefore, they would have to wait after their first break together to get together again after 840 minutes

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3 years ago
What is 5/2y +17= -2/3y -21
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19/6y+17=-21
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3 years ago
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The position function of a particle moving along a coordinate line is given, wheresis in feet andtis in seconds.
Sati [7]

Answer:

a) s = (4-t)/(t^2+4)^2,   a(t) = (2t^3-24t)/(t^2+4)^3

b) s = 0.2ft, v = 0.12 ft/s,  a = -0.176 ft/s^2

c) t = 2s

d) slowing down for t < 2, speeding up for t > 2

e) 0.327 ft

Step-by-step explanation:

The position function of a particle is given by:

s(t)=\frac{t}{t^2+4},\ \ \ t\geq  0   (1)

a) The velocity function is the derivative, in time, of the position function:

v(t)=\frac{ds}{dt}=\frac{(1)(t^2+4)-t(2t)}{(t^2+4)^2}=\frac{4-t^2}{(t^2+4)^2}   (2)

The acceleration is the derivative of the velocity:

a(t)=\frac{dv}{dt}=\frac{(-2t)(t^2+4t)^2-(4-t^2)2(t^2+4)(2t)}{(t^2+4)^4}\\\\a(t)=\frac{(-2t)(t^2+4)-4t(4-t^2)}{(t^2+4)^3}=\frac{2t^3-24t}{(t^2+4)^3} (3)

b) For t = 1 you have:

s(1)=\frac{1}{1+4}=0.2\ ft\\\\v(1)=\frac{4-1}{(1+4)^2}=0.12\frac{ft}{s}\\\\a(1)=\frac{2-24}{(1+4)^3}=-0.176\frac{ft}{s^2}

c) The particle stops for v(t)=0. Then you equal equation (2) to zero ans solve the equation for t:

v(t)=\frac{4-t^2}{(t^2+4)^2}=0\\\\4-t^2=0\\\\t=2

For t = 2s the particle stops.

d) The second derivative evaluated in t=2 give us the concavity of the position function.

\frac{d^2s}{dt^2}=a(2)=\frac{2(2)^3-24(2)}{(2^2+4)^3}=-0.062

Then, the concavity of the position function is negative. For t=2 there is a maximum. Before t=2 the particle is slowing down and after t=2 the particle is speeding up.

e) Due to particle goes and come back. You first calculate s for t=2, then calculate for t=5.

s(2)=\frac{2}{2^2+4}=0.25\ ft

s(5)=\frac{5}{5^2+4}=0.172\ ft

The particle travels 0.25 in the first 2 seconds. In the following three second the particle comes back to the 0.172\ ft. Then, in the second trajectory the particle travels:

0.25 - 0.127 = 0.077 ft

The total distance is the sum of the distance of the two trajectories:

s_total = 0.25 ft + 0.077 ft = 0.327 ft

6 0
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120, 160, 200 <br><br>Find the gcf<br><br>please show me the work.
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120|\fbox{2}\\..60|\fbox{2}\\..30|\fbox{2}\\..15|3\\....5|\fbox{5}\\....1|\\-------------\\160|\fbox{2}\\..80|\fbox{2}\\..40|\fbox{2}\\..20|2\\..10|2\\....5|\fbox{5}\\....1|
----------\\200|\fbox{2}\\100|\fbox{2}\\..50|\fbox{2}\\..25|\fbox{5}\\....5|5\\....1|


GCF(120;\ 160;\ 200)=2\times2\times2\times5=40
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3 years ago
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