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A point PPP moves along the x-axis. Its position after ttt seconds is given by x=2t3−21t2+60tx=2t^3-21t^2+60tx=2t3−21t2+60t meters.

1. Find the velocity and the acceleration.

v(t)=

a(t)=

I don't know

A rock is thrown upwards and its position (in meters) above the ground is given by s(t)=−5t2+20t+2s\left(t\right)=-5t^2+20t+2s(t)=−5t2+20t+2, where ttt is in seconds.
a) Find i) the maximum height the rock reaches, and ii) when it reaches this height.
b) Find the velocity of the rock when it is 22m above the ground.
c) Determine the velocity of the rock when it hits the ground

a) i) Enter the exact value of the maximum height of the rock (in metres)

a) ii) Enter the exact time in which this max height was reached (in seconds)

b) Enter the exact velocity of the rock at this point (in m/s)

c) Enter the velocity of the rock when it hits the ground (in m/s), rounded to 2 decimal place

I don't know

 x(t)=6t3+3t2−3,x(t)=6t^3+3t^2-3,x(t)=6t3+3t2−3, Find velocity and acceleration at t=2?

velocity

acceleration

I don't know

Two cities, Growth and Decay, have populations that are respectively increasing and decreasing at different rates that are proportional to their current population. Growth’s population is now 4 millions and was 2 millions 10 years ago. Decay’s population is now 6 millions and was 8 millions 10 years ago. 

In how many years from now will Growth and Decay’s populations be equal? 

10ln⁡(4)ln⁡(2)+ln⁡(6/8)−10\frac{10\ln(4)}{\ln(2)+\ln(6/8)}-10ln(2)+ln(6/8)10ln(4)​−10

10ln⁡(4)ln⁡(2)−ln⁡(6/8)+10\frac{10\ln(4)}{\ln(2)-\ln(6/8)}+10ln(2)−ln(6/8)10ln(4)​+10

10ln⁡(4)ln⁡(2)−ln⁡(6/8)−10\frac{10\ln(4)}{\ln(2)-\ln(6/8)}-10ln(2)−ln(6/8)10ln(4)​−10

10ln⁡(4)ln⁡(2)+ln⁡(6/8)+10\frac{10\ln(4)}{\ln(2)+\ln(6/8)}+10ln(2)+ln(6/8)10ln(4)​+10

I don't know

 A bacteria culture grows exponentially. After one day, there are  40×10440\times10^440×104  bacteria. After 5 days, there are 40×10640\times10^640×106  bacteria. 

 Find the population of bacteria after tttdays. 

4×104e(1/2)ln⁡(10)e(1/2)ln⁡(10)t\frac{4\times10^4}{e^{(1/2)\ln(10)}}e^{(1/2)\ln(10)t}e(1/2)ln(10)4×104​e(1/2)ln(10)t

40×102e(1/2)ln⁡(10)e(1/2)ln⁡(10)t\frac{40\times10^2}{e^{(1/2)\ln(10)}}e^{(1/2)\ln(10)t}e(1/2)ln(10)40×102​e(1/2)ln(10)t

40×104e(1/2)ln⁡(10)e(1/2)ln⁡(10)t\frac{40\times10^4}{e^{(1/2)\ln(10)}}e^{(1/2)\ln(10)t}e(1/2)ln(10)40×104​e(1/2)ln(10)t

4×102e(1/2)ln⁡(10)e(1/2)ln⁡(10)t\frac{4\times10^2}{e^{(1/2)\ln(10)}}e^{(1/2)\ln(10)t}e(1/2)ln(10)4×102​e(1/2)ln(10)t

I don't know

 A sample of radioactive material decayed to 15% of its original amount after 10 years. What is the half-life of the material?  

10ln⁡(1/2)ln⁡(0.15)10\frac{\ln(1/2)}{\ln(0.15)}10ln(0.15)ln(1/2)​

100ln⁡(1/2)ln⁡(0.15)100\frac{\ln(1/2)}{\ln(0.15)}100ln(0.15)ln(1/2)​

10ln⁡(0.15)ln⁡(1/2)10\frac{\ln(0.15)}{\ln(1/2)}10ln(1/2)ln(0.15)​

100ln⁡(0.15)ln⁡(1/2)100\frac{\ln(0.15)}{\ln(1/2)}100ln(1/2)ln(0.15)​

I don't know

 A sample of radioactive material decayed to 15% of its original amount after 10 years. What percentage is remaining after another 5 years? 

10e(1/100)ln⁡(0.15)1510e^{(1/100)\ln(0.15)15}10e(1/100)ln(0.15)15

100e(1/100)ln⁡(0.15)15100e^{(1/100)\ln(0.15)15}100e(1/100)ln(0.15)15

10e(1/10)ln⁡(0.15)1510e^{(1/10)\ln(0.15)15}10e(1/10)ln(0.15)15

100e(1/10)ln⁡(0.15)15100e^{(1/10)\ln(0.15)15}100e(1/10)ln(0.15)15

I don't know

Let’s learn how to cook pulled pork. A pork shoulder is removed from a refrigerator at 0∘C0^{\circ}C0∘C and placed into an already-hot slow cooker. The cooker initially cooks the pork for 2 hours on HIGH (150∘C)(150^{\circ}C)(150∘C), after which it is switched to LOW (100∘C)(100^{\circ}C)(100∘C) for another 8 hours. Assume the temperature of the cooker switches instantaneously and the pork follows Newton’s law of heating.

If the pork is at 30∘C30^{\circ}C30∘C after 1 hour, what is its internal temperature when the cooker is switched to LOW?

Answer

I don't know