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The answer is 30m. Lets assume that the ball is dropped from a height of h. The ball will come down and go up, so in the first bounce it covers h+h/2 distance. The second bounce, it is h/2+h/4, the third it will be h/4+h/8 and so on. The total distance covered would thus be h+h/2+h/2+h/4+h/4+h/8+h/8+........... = h+h+h/2+h/4+h/8+........... (summing up adjacent values in pairs) = 2h+h*(1/2+1/4+1/8+.............) = 3h (by the geometric series formula, 1/2+1/4+1/8+.....=1) Hence taking, h=10m in this case, the answer would be 10*3= 30m
Each flip can land in 2 ways. Total number of possible ways that 5 flips can land is2 x 2 x 2 x 2 x 2 = 32 different ways.There are only 5 different ways to get exactly 3 heads in a row in 5 flips.They are:H H H T TH H H T HT H H H TT T H H HH T H H H.The probabilty is 5/32 = 15.625 percent.====================================================That's if you mean exactly 3 heads in a row ... no less and no more.What if you asked for "at least 3 heads in a row" ?There are 8 ways to do that:The original five ways to get 3 in a row . . .H H H T TH H H T HT H H H TT T H H HH T H H HPlus 2 ways to get 4 in a row . . .T H H H HH H H H TPlus a way to get 5 in a row . . .H H H H H.The probability of one of these outcomes is 8/32 = 25 percent.
A = (bh)/2 = [(4h - 2)h]/2 = (2h - 1)h = 22h - h Restriction: 2h - 1 > 0 implies h > 1/2, since the area cannot be zero or negative.
d = r/2 = 1.3656 V = pi x (d/2)2 x h h = d/2 and V = 2 (d/2)2 times h = 2/pi 2/pi = 0.636619772
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
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By Pythagoras h^2 = a^2 + b^2 Substituting h^2 = 12^(2) + 5^(2) h^2 = 144 + 25 h^2 = 169 h = sqrt(169) h = 13
d/dh(h^-1) = -1(h^-2) = -(h^-2)
Factor 2h−2 2h−2 =2(h−1) Answer: 2(h−1)
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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], ...
V = pi r^2*H & V = pi (4r)^2*h Equate pi r^(2)H= pi (4r)^2 h pi cancel down r^(2)H = (4r)^2h r^(2)*H = 16r^(2)*h 'r^(2) cancels down H= 16h h = H/16 This means is you increase the radius by '4 times' , then you reduce the height(H) by '16 times' in order to maintain the same volume.
The length of the hypotenuse is: 583.1
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].
[Hypotenuse]2 = 92 + 152 = 81 + 225 = 306 So hypotenuse = sqrt(306) = 17.49 to 2 dp