16t2 - 37t + 20 = 0 Using the quadratic formula, t = [37 +/- sqrt(372 - 4*16*20)]/(2*16) = [37 +/- sqrt(89)] / 32 ie t = 0.861438 or t = 1.451062
The bounds of integration are 10 and 20. The function that we are integrating is Q(t)=4(.96t)=3.84t. So the average value of Q(t) from 10 to 20 is equal to [1/(20-10)]*the integral from 10 to 20 of 3.84t dt. Simplifying, we get .384*the integral from 10 to 20 of t dt. Integrating, we get that the average value = .384(20^2 - 10^2)/2 = .196*(400 - 100) = .196 * 300 = 58.8. The average value in question is exactly 58.8 grams.
Pseudocode is generally a very loosely defined concept. Various ways you can show your statement: if y = 20 then x = 0 if( y == 20 ) x = 0 if y is 20 then set x to 0
t=0:.02:20 > num1=[1] > num2=[1 0] > denum=[.2 .2 6] > sys1=tf(num1,denum) > sys2=tf(num2, denum) > xt=impulse(sys1,t) > xdott=impulse(sys2,t) > plot(t,xt,'r',t,xdott,'b') >
That depends upon what the object has done for the 20 seconds since t = 0 seconds.
2 to the power 0 or 20 is equal to 1
There are infinitely many quadratics that can equal 20.For example, given any positive number, a, a*x^2 + 20 will equal 20 for x= 0.
16t2 - 37t + 20 = 0 Using the quadratic formula, t = [37 +/- sqrt(372 - 4*16*20)]/(2*16) = [37 +/- sqrt(89)] / 32 ie t = 0.861438 or t = 1.451062
6
The bounds of integration are 10 and 20. The function that we are integrating is Q(t)=4(.96t)=3.84t. So the average value of Q(t) from 10 to 20 is equal to [1/(20-10)]*the integral from 10 to 20 of 3.84t dt. Simplifying, we get .384*the integral from 10 to 20 of t dt. Integrating, we get that the average value = .384(20^2 - 10^2)/2 = .196*(400 - 100) = .196 * 300 = 58.8. The average value in question is exactly 58.8 grams.
r=0,Tr-r = 0 = r(T-1), since T != 1, then T-1 is non zero so r must be zero.
Pseudocode is generally a very loosely defined concept. Various ways you can show your statement: if y = 20 then x = 0 if( y == 20 ) x = 0 if y is 20 then set x to 0
u= 0 m/s v= 200 m/s m= 20 kg F= 100 N F= (mv - mu)/t 100 = (20*200 - 20*0)/t 100 = 4000/t t= 4000/100 t= 40 s hope it helps cheers ^_^
v=u+(a*t) 0=500+(a*20) (0-500)/20=a a=-25(km/h)/s
That depends upon what the object has done for the 20 seconds since t = 0 seconds.
t=0:.02:20 > num1=[1] > num2=[1 0] > denum=[.2 .2 6] > sys1=tf(num1,denum) > sys2=tf(num2, denum) > xt=impulse(sys1,t) > xdott=impulse(sys2,t) > plot(t,xt,'r',t,xdott,'b') >