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Q: What is x h divided by x h?

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The associative property.

Apply the reciprocal rule: If f(x) = 1/h(x) then f'(x) = -h'(x)/(h(x))^2

Volume = height x pi x diameter squared divided by 4 V = H x pi x D^2/4 D = sqrt ((4x V) / (H x pi))

f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2

f'[x] = lim(h->0) (f[x+h]-f[x])/h lim(h->0) (sin[x+h]-sin[x])/h By angle-addition formula, we have: lim(h->0) (sin[x]cos[h]+sin[h]cos[x]-sin[x])/h lim(h->0) (sin[x]cos[h]-sin[x])/h + lim(h->0) (sin[h]cos[x])/h sin[x]*lim(h->0) (cos[h]-1)/h + cos[x]*lim(h->0) sin[h]/h In a calculus class, it is shown that: lim(h->0) (cos[h]-1)/h = 0 and that lim(h->0) sin[h]/h is 1. So, sin[x]*lim(h->0) (cos[h]-1)/h + cos[x]*lim(h->0) sin[h]/h becomes sin[x]*0 + cos[x]*1 cos[x] So, if f[x] = sin[x], f'[x] = cos[x]

Related questions

The associative property.

Apply the reciprocal rule: If f(x) = 1/h(x) then f'(x) = -h'(x)/(h(x))^2

Pi x R sq x H x 7.48If in inches Pi x Radius Sq x H divided by 2.31

lim(h→0) (sin x cos h + cos x sin h - sin x)/h As h tends to 0, both the numerator and the denominator have limit zero. Thus, the quotient is indeterminate at 0 and of the form 0/0. Therefore, we apply l'Hopital's Rule and the limit equals: lim(h→0) (sin x cos h + cos x sin h - sin x)/h = lim(h→0) (sin x cos h + cos x sin h - sin x)'/h' = lim(h→0) [[(cos x)(cos h) + (sin x)(-sin h)] + [(-sin x)(sin h) + (cos x)(cos h)] - cos x]]/0 = cosx/0 = ∞

Volume = height x pi x diameter squared divided by 4 V = H x pi x D^2/4 D = sqrt ((4x V) / (H x pi))

Since the Volume=Length x Width x Height (or V=L x W x H), then L= V / (W x H). Length=Volume divided by <Width x Height>

f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2f(x) = x/2Then the differential is lim h->0 [f(x+h) - f(x)]/h= lim h->0 [(x+h)/2 - x/2]/h= lim h->0 [h/2]/h= lim h->0 [1/2] = 1/2

f'[x] = lim(h->0) (f[x+h]-f[x])/h lim(h->0) (sin[x+h]-sin[x])/h By angle-addition formula, we have: lim(h->0) (sin[x]cos[h]+sin[h]cos[x]-sin[x])/h lim(h->0) (sin[x]cos[h]-sin[x])/h + lim(h->0) (sin[h]cos[x])/h sin[x]*lim(h->0) (cos[h]-1)/h + cos[x]*lim(h->0) sin[h]/h In a calculus class, it is shown that: lim(h->0) (cos[h]-1)/h = 0 and that lim(h->0) sin[h]/h is 1. So, sin[x]*lim(h->0) (cos[h]-1)/h + cos[x]*lim(h->0) sin[h]/h becomes sin[x]*0 + cos[x]*1 cos[x] So, if f[x] = sin[x], f'[x] = cos[x]

2 divided by x-12-x divided by x-2 is -1/(6(x-2)).

The formula I think is this: l x w x h /2l=lengthw=widthh=height/ =Divided byx =multiplied by2= 2

-x divided by -5 is x/5.

Undefined: You cannot divide by zero