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What is the factor of 30 d to the 5 power?
2 x 3 x 5 x d x d x d x d x d = 30d5
What is the factor of the monomial 30d5?
2 x 3 x 5 x d x d x d x d x d = 30d5
What is 4d x d x d?
4d x d x d = 4d3
Where can you buy Alex rider the Gemini project?
Well for one you cant buy Alex Rider the Gemini Project because it doesn't exist what your after is Alex Rider Point Blanc and as for that you can buy it in many places like amazon or got to the libary and get it :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :D :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :p :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :o :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :b :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :x :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :) :)
How do you prove that the derivative of sec x is equal to sec x tan x?
Show that sec'x = d/dx (sec x) = sec x tan x. First, take note that sec x = 1/cos x; d sin x = cos x dx; d cos x = -sin x dx; and d log u = du/u. From the last, we have du = u d log u. Then, letting u = sec x, we have, d sec x = sec x d log sec x; and d log sec x = d log ( 1 / cos x ) = -d log cos x = d ( -cos x ) / cos x = sin x dx / cos x = tan x dx. Thence, d sec x = sec x tan x dx, and sec' x = sec x tan x, which is what we set out to show.
What is the derivative of ddx(a2x)?
Assuming you mean what is the value of the derivative d/dx(a²x), then: d/dx(a²x) = a² The derivative (with respect to x) of d/dx(a²x) = d/dx(d/dx(a²x)) = d/dx(a²) = 0.
What is the derivative of secxtanx?
d/dx(uv)=u*dv/dx+v*du/dxd/dx(secxtanx)=secx*[d/dx(tanx)]+tanx*[d/dx(secx)]-The derivative of tanx is:d/dx(tan u)=[sec(u)]2*d/dx(u)d/dx(tan x)=[sec(x)]2*d/dx(x)d/dx(tan x)=[sec(x)]2*(1)d/dx(tan x)=(sec(x))2=sec2(x)-The derivative of secx is:d/dx(sec u)=[sec(u)tan(u)]*d/dx(u)d/dx(sec x)=[sec(x)tan(x)]*d/dx(x)d/dx(sec x)=[sec(x)tan(x)]*(1)d/dx(sec x)=sec(x)tan(x)d/dx(secxtanx)=secx*[sec2(x)]+tanx*[sec(x)tan(x)]d/dx(secxtanx)=sec3(x)+sec(x)tan2(x)
Formula to calculate v-belt length?
The answer is: L = pi x (D + d)/2 + 2 x ( C x Cos(a) + a x (D-d)/2) where a = arcsin(D-d)/(2 x C) in radians. Where C is the center distance, D is the large pulley diameter, and d is the small pulley diameter.
What does d mean in 1 x c x 3 equals d?
d=3c
What is the derivative of f plus g of 3 and f times g of 3 given that f of 3 equals 5 d dx f of 3 equals 1.1 g of 3 equals -4 d dx g of 3 equals 7 Also please explain QUICK THANK YOU?
d/dx [f(x) + g(x)] = d/dx [f(x)] + d/dx [g(x)] or f'(x) + g'(x) when x = 3, d/dx [f(x) + g(x)] = f'(3) + g'(3) = 1.1 + 7 = 8.1 d/dx [f(x)*g(x)] = f(x)*d/dx[g(x)] + d/dx[f(x)]*g(x) when x = 3, d/dx [f(x)*g(x)] = f(3)*g'(3) + f'(3)*g(3) = 5*7 + 1.1*(-4) = 35 - 4.4 = 31.1
What is derivative of 1 - x?
d/dx (1-x)=d/dx (1)-d/dx(x)=0-1=-1
How do you differentiate x sinx with respect to x?
You should apply the chain rule d/dx(x.sin x) = x * d/dx(sin x) + sin x * d/dx(x) = x * cos x + sin x * 1 = x.cos x + sin x
