cidyah
cidyah
About
- Username
- cidyah
- Joined
- Visits
- 0
- Last Active
- Roles
- Member
Comments
-
I haven't solved it. But this is a graph and the area they are looking for. 10 cos(theta)=5 cos(theta)=5/10 = 1/2 theta = pi/3, 5pi/3 I have enclosed a link that may be of use. Click here for the area picture: http://s1169.photobucket.com…
-
a) f(t) = t-sin(t) Graph of f(t) shows it is strictly increasing http://www.wolframalpha.com/input/?i=graph+t-sin(t... f'(t) = 1- cos(t) f''(t) = sin(t) = 0 t = n pi , where n is an integer (inflection points)
-
false.
-
a) f(x) = 3x^4 + 4x^3 - 12x^2 + 10 f'(x)=12x^3+12x^2-24x f'(x)=12x(x^2+x-2)=0 x=0 x^2+x-2=0 (x+2)(x-1)=0 x=1, x=-2 The critical numbers are x=0, x=1, x=-2 b) Consider the intervals (-∞,-2),(-2,0),(0,1),(1,∞) choose any one point from…
-
a) f(x) = 3x^4 + 4x^3 - 12x^2 + 10 f'(x)=12x^3+12x^2-24x f'(x)=12x(x^2+x-2)=0 x=0 x^2+x-2=0 (x+2)(x-1)=0 x=1, x=-2 The critical numbers are x=0, x=1, x=-2 b) Consider the intervals (-∞,-2),(-2,0),(0,1),(1,∞) choose any one point from…
-
substitute x=0, we have 0/0 Apply L'Hopital's rule lim x-->0 (1-cos 5x) / 7x^2 = lim x-->0 5 sin 5x / 14 x = lim x-->0 25 cos 5x / 14 = 25/14
-
(1-cos^2 theta)(1+cot^2 theta)= sin^2(theta) csc^2(theta) = sin^2(theta) / sin^2(theta) = 1
-
v = e^(6xe^y) ∂v/∂x = e^(6xe^y) (6e^y) ∂v/∂x = 6 e^y e^(6xe^y) ----------(1) ∂v/∂y = e^(6xe^y) 6xe^y ∂v/∂y = 6x e^y e^(6xe^y) ------------(2) xx = ∂^2v/∂x^2 = differentiate (1) with respect to x only (y is a constant) = 6e^y e^(6xe^y) …
-
v = e^(6xe^y) ∂v/∂x = e^(6xe^y) (6e^y) ∂v/∂x = 6 e^y e^(6xe^y) ----------(1) ∂v/∂y = e^(6xe^y) 6xe^y ∂v/∂y = 6x e^y e^(6xe^y) ------------(2) xx = ∂^2v/∂x^2 = differentiate (1) with respect to x only (y is a constant) = 6e^y e^(6xe^y) …
-
Did you mean the partial derivative with respect to y? If so: F(x,y) = y^3 (sin (y^2 x) ln (4y+x) differentiate with respect to y only treating x as a constant. Fy(x,y) = (3y^2) (sin (y^2 x) ln (4y+x) + y^3 cos ((y^2) x) (2yx) ln (4y+x) + y^3…
-
Did you mean the partial derivative with respect to y? If so: F(x,y) = y^3 (sin (y^2 x) ln (4y+x) differentiate with respect to y only treating x as a constant. Fy(x,y) = (3y^2) (sin (y^2 x) ln (4y+x) + y^3 cos ((y^2) x) (2yx) ln (4y+x) + y^3…
-
cos(70) = x/8 x = 8 cos(70) x = 2.74 m sin(70) = y/8 y = 8 sin(70) y = 7.52 m https://gyazo.com/a718371c0e5348bcf3d0f6fd64d79316
-
-1 and 6 are two solutions by trial and error. The other two roots may be complex.
-
-1 and 6 are two solutions by trial and error. The other two roots may be complex.
-
f(x) = 7x - 2x ln (x) f'(x) = 7 - 2 ln(x) - 2x / x f'(x) = 7 - 2 ln (x) -2 = 0 -2 ln (x) = -5 ln(x) = 5/2 x=e^(5/2) is a critical point f''(x) = -2/x < 0 when x=e^5, so f(x) has a local global maximum when x=e^(5/2) The global maximum i…
-
p(x) = sqrt(f(x)) p(x)= [f(x)]^(1/2) p'(x)= (1/2) [f(x)]^(-1/2)] f'(x) p'(x)= 1/2[f(x)]^(1/2) f'(x) p'(x) = f'(x)/2sqrt((f(x))
-
p(x) = sqrt(f(x)) p(x)= [f(x)]^(1/2) p'(x)= (1/2) [f(x)]^(-1/2)] f'(x) p'(x)= 1/2[f(x)]^(1/2) f'(x) p'(x) = f'(x)/2sqrt((f(x))
-
(a) Calculate the price that the town's fishery should charge for tuna in order to produce a demand of 400 pounds of tuna per month. p = 24,000/q^1.5 Substitute q=400 into the equation p = 24,000 /400^1.5 p = 24,000 /8,000 = $3,000 per lb b…
-
(a) Calculate the price that the town's fishery should charge for tuna in order to produce a demand of 400 pounds of tuna per month. p = 24,000/q^1.5 Substitute q=400 into the equation p = 24,000 /400^1.5 p = 24,000 /8,000 = $3,000 per lb b…
-
lim x-->∞ 4x^2 /(x^2-4) = lim x-->∞ 4x^2 / (x^2 (1-4/x^2) = lim x-->∞ 4 / (1-4/x^2) As x approaches ∞, 4/x^2 approaches 0 lim x-->∞ 4 /1 = 4 y= 4 is the horizontal asymptote
-
a) x(t)= e^ (-t) sin t x'(t) = -e^(-t) sin t + e^(-t) cos t = 0 e^(-t) ( cos t - sin t) = 0 cos t = sin t divide both sides by sin t tan t = 1 t = PI/4 t = 5PI/4 x''(t) = e^(-t) sin t -e^(-t) cos t - e^(-t) cos t - e^(-t) sin t = -2 e^(-t…
-
a) x(t)= e^ (-t) sin t x'(t) = -e^(-t) sin t + e^(-t) cos t = 0 e^(-t) ( cos t - sin t) = 0 cos t = sin t divide both sides by sin t tan t = 1 t = PI/4 t = 5PI/4 x''(t) = e^(-t) sin t -e^(-t) cos t - e^(-t) cos t - e^(-t) sin t = -2 e^(-t…
-
lim x-->∞ 4x^2 /(x^2-4) = lim x-->∞ 4x^2 / (x^2 (1-4/x^2) = lim x-->∞ 4 / (1-4/x^2) As x approaches ∞, 4/x^2 approaches 0 lim x-->∞ 4 /1 = 4 y= 4 is the horizontal asymptote
-
lim x-->∞ 4x^2 /(x^2-4) = lim x-->∞ 4x^2 / (x^2 (1-4/x^2) = lim x-->∞ 4 / (1-4/x^2) As x approaches ∞, 4/x^2 approaches 0 lim x-->∞ 4 /1 = 4 y= 4 is the horizontal asymptote
-
(a) Calculate the price that the town's fishery should charge for tuna in order to produce a demand of 400 pounds of tuna per month. p = 24,000/q^1.5 Substitute q=400 into the equation p = 24,000 /400^1.5 p = 24,000 /8,000 = $3,000 per lb b…
-
p(x) = sqrt(f(x)) p(x)= [f(x)]^(1/2) p'(x)= (1/2) [f(x)]^(-1/2)] f'(x) p'(x)= 1/2[f(x)]^(1/2) f'(x) p'(x) = f'(x)/2sqrt((f(x))
-
f(x) = 7x - 2x ln (x) f'(x) = 7 - 2 ln(x) - 2x / x f'(x) = 7 - 2 ln (x) -2 = 0 -2 ln (x) = -5 ln(x) = 5/2 x=e^(5/2) is a critical point f''(x) = -2/x < 0 when x=e^5, so f(x) has a local global maximum when x=e^(5/2) The global maximum i…
-
f(x) = 7x - 2x ln (x) f'(x) = 7 - 2 ln(x) - 2x / x f'(x) = 7 - 2 ln (x) -2 = 0 -2 ln (x) = -5 ln(x) = 5/2 x=e^(5/2) is a critical point f''(x) = -2/x < 0 when x=e^5, so f(x) has a local global maximum when x=e^(5/2) The global maximum i…
-
-1 and 6 are two solutions by trial and error. The other two roots may be complex.
-
-1 and 6 are two solutions by trial and error. The other two roots may be complex.