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3 years ago

Determine the m m m m m

Mathematics

1 answer:

iVinArrow [24]3 years ago

5 0

Answer:

A,D,E,G=83 B,C,F=97

Step-by-step explanation:

You might be interested in

The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s

Norma-Jean [14]

The positions when the particle reverses direction are:

$s(t_1)=55ft\\\\s(t_2)=28ft$

The acceleraton of the paticle when reverses direction is:

$a(t_1)=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=18\frac{ft}{s^{2}}$

Why?

To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

Also, when the particle reverses its direction, the velocity is equal to 0.

We are given the following function:

$s(t)=2t^{3}-21t^{2}+60t+3$

So,

- Derivating to get the velocity function, we have:

$v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60$

Now, making the function equal to 0, to find the times when the particle reversed its direction, we have:

$v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s$

We know that the particle reversed its direction two times.

- Derivating the velocity function to find the acceleration function, we have:

$a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42$

Now, substituting the times to calculate the accelerations, we have:

$a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}$

Now, substitutitng the times to calculate the positions, we have:

$s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft$

Have a nice day!

3 0

3 years ago

Can someone please help me?

Maurinko [17]

Yessssssssssssssssssssssssss

5 0

3 years ago

2x + y = 12

Arlecino [84]

Answer:

Whats the question?

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

kelly buys shoes with all the of her money. she only has $150. one pair of shoes is for her and they cost $40. the other 2 pairs

shusha [124]

$150 - $40=110

$110/2=55

6 0

3 years ago

2. A solar lease customer built up an excess of 6,500 kilowatt hours (kwh) during the summer using his solar

tia_tia [17]

9514 1404 393

Answer:

a) E = 6500 -50d

b) 5000 kWh

c) the excess will last only 130 days, not enough for 5 months

Step-by-step explanation:

Given:

starting excess (E): 6500 kWh

usage: 50 kWh/day (d)

Find:

a) E(d)

b) E(30)

c) E(150)

Solution:

a) The exces is linearly decreasing with the number of days, so we have ...

E(d) = 6500 -50d

b) After 30 days, the excess remaining is ...

E(30) = 6500 -50(30) = 5000 . . . . kWh after 30 days

c) After 150 days, the excess remaining would be ...

E(150) = 6500 -50(150) = 6500 -7500 = -1000 . . . . 150 days is beyond the capacity of the system

The supply is not enough to last for 5 months.

3 0

2 years ago