g(t) = 2/tThe function is the same as writing g(t) = 2 t-1,and that's not too difficult to differentiate:g'(t) = -2 t-2g'(1/2) = -2 (1/2)-2 = -2 (4) = -8
divided by what? m will be equals to t divided by v
T=theta so that it will not look so messy. g(T)=TcscT To find the first derivative, you must use the product rule. Product rule is derivative of the first times the second, plus the first times the derivative of the second, which will give you: g'(T)=0xcscT + Tx-cscTcotT, which simplifies: g'(T)= -cscTxcotT Now, take the derivative of that to get the second derivatice. In order to do that, you have to do the product rule again. g"(T)=(cscTcotT)cotT + -cscT(-csc^2T) {that's csc squared} which simplifies: g"(T)= cscTcot^2(T) + csc^3 (T)
(vf-vi)/ t is ?
Yes, V (velocity) = d (distance) divided by t (time).
g(t) = 2/tThe function is the same as writing g(t) = 2 t-1,and that's not too difficult to differentiate:g'(t) = -2 t-2g'(1/2) = -2 (1/2)-2 = -2 (4) = -8
divided by what? m will be equals to t divided by v
(vf-vi)/ t is ?
T=theta so that it will not look so messy. g(T)=TcscT To find the first derivative, you must use the product rule. Product rule is derivative of the first times the second, plus the first times the derivative of the second, which will give you: g'(T)=0xcscT + Tx-cscTcotT, which simplifies: g'(T)= -cscTxcotT Now, take the derivative of that to get the second derivatice. In order to do that, you have to do the product rule again. g"(T)=(cscTcotT)cotT + -cscT(-csc^2T) {that's csc squared} which simplifies: g"(T)= cscTcot^2(T) + csc^3 (T)
Yes, V (velocity) = d (distance) divided by t (time).
T divided by 30 30 times another number equals T
d=r/t (D equals r divided on t
0.8333
1.5
no it equals 20, yes it equals 2 you illiterate piece of sh*t
If 2.3 times t = 90.... 90 divided by 2.3 = 39.1 therefor t=39.1
9 + t/12 = -3 t/12 = -12 t = -144