2 Humps on a Bactrian Camel
2 Hands on a Clock
2 Hulls on a Catamaran
2 halves in a whole
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
-1
d/dh(h^-1) = -1(h^-2) = -(h^-2)
H-H or H:H depending on what you and your professors prefer.
Factor 2h−2 2h−2 =2(h−1) Answer: 2(h−1)
15
Slice the bowl horizontally into circles, then integrate the area of the circles. The area of each circle is (pi * r^2). The height of each slice is dh. The 1st (bottom) circle is r=0. The r^2 of each circle-slice is (2*A*h-h^2), where A is the spherical radius, and h is the variable height of any given slice. At the top of the water level, (r^2=2*A*H-H^2). Integrate the area over the interval h=0->H as follows: V=pi * integral[(2*A*h - h^2) dh]; h=0->H to yield V=pi * (2*A*h^2 / 2 - h^3 / 3); h=0->H V=pi * (A*H^2 - H^3 / 3). As a check, plug the full diameter (2*A) in for H. If you did the integration correctly, you will get the full volume of the sphere, (4/3 * pi * A^3).
There are 2 outcomes for the dime (H or T), 2 for the penny (H or T) and 6 for the die (1,2,3,4,5,6). In all, there are 2*2*6 = 24 outcomes. Some of them are given below in the pattern: dime, penny, die. The rest are easy to generate. [H,H,1], [H,H,2], ... , [H,H,6], [H,T,1], [H,T,2], ... [T,H,1], ... [T,T,1], ...
Being a Right- Angled triangle, apply Pythagoras. h^2 = a^2 + b^2 h^2 = 500^2 + 300^2 h^2 = 250000 + 90000 h^2 = 340000 h = sqrt(340000) h = 583.095...
multiplication is point to point and convolustion is point to multi-point ex multiplication-- s[n]=x[n].h[n] s[0]=[x[0].h[0] s[1]=[x[1].h[1] s[2]=[x[2].h[2] . . . .. s[n-1]=[x[n-1].h[n-1] convollustion s[n]=x[n]*h[n] s[0]=[x[0].h[0]+x[0].h[1]+x[0].h[2]+.......+x[0].h[n-1] s[1]=[x[1].h[0]+x[1].h[1]+x[1].h[2]+.......+x[1].h[n-1] s[2]=[x[2].h[2]+x[2].h[1]+x[2].h[2]+.......+x[2].h[n-1] . . . s[n-1]=[x[n-1].h[0]+x[n-1].h[1]+x[n-1].h[2]+.......+x[n-1].h[n-1].
Use Pythagoras h^2 = a^2 + b^2 Substitute h^2 9^2 + 15^2 h^2 = 81 + 225 h^2 = 306 h = sqrt(306) h = 17.429...
H-1, H-2, H-3 and H-201.