Its height (h) is side (s) times sin 60 orsqrt(3)/2 times side = .866s h = .866s s = h/.866 = 2/sqrt(3) s = 1.154 h perimeter = s + s + s = 3s = 3 x 2/sqrt(3) = 3.46s
Re-arrange this equation: L=length W=width H=height S=surface area S = 2(LW+HW+HL) S/2 = LW+HW+HL (S/2) - LW = HW+HL (S/2) - LW = H(W+L) ((S/2) - LW)/(W+l) = H
7.7811/sqrt(2)=7.78If the length of a leg of a 45-45-90 right triangle is s, then the hypotenuse, h, is s*sqrt(2).h=s*sqrt(2) --> s=h/sqrt(2) [This is just a special case of the Pythagorean theory.]
72000
The solution is as follows. First find the final velocity. Second use the final velocity to find the time. (final velocity squared) = (initial velocity squared) + 2gh (vf)^2=(vi)^2+2gh (vf)^2=(30ft/s)^2+2(32ft/s^2)(80ft) (vf)^2=900(ft/s)^2+5120(ft/s)^2=6028(ft/s)^2 taking the square root yields vf=77.59(ft/s) The final velocity and the time are related by the equation vf=vi+gt, rearranging t=(vf-vi)/g t=[77.59(ft/s)-30(ft/s)]/32(ft/s^2=47.59(ft/s)/32ft/s^2=1.49s You can check the answer as follows: h=vi(t)+(1/2)gt^2 h=30(ft/s)(1.49s)+(1/2)(32(ft/s^2)(1.49s)^2 h=44.7ft+16(ft/s^2)[2.22s^2) h=44.7ft+35.7ft=80.2ft which is equivalent to the 80 ft in the original problem except for rounding errors. Note: the notation here is awkward but I haven't found an equation editor iw Wiki. h^2 means h squared ft/s^2 means feet per second squared
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].
Its height (h) is side (s) times sin 60 orsqrt(3)/2 times side = .866s h = .866s s = h/.866 = 2/sqrt(3) s = 1.154 h perimeter = s + s + s = 3s = 3 x 2/sqrt(3) = 3.46s
2 hemispheres in a sphere.
nobody because they thought it was a piece of S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S S H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H H I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I I T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T T
S = 2 pi r h + 2 pi r2S - 2 pi r2 = 2 pi r hh = (S - 2 pi r2) / (2 pi r)
The answer depends on the information that you have.If you know a side (s) and an angle (A), then area = s^2*sin(A)If you know the side (s) and height (h) then area = 1/2*s*h
This is ridiculous
v = 45 mph t = 2 minutes = 1/30 h S = ? S = vt S = 45 miles/h x (1/30) h S = 1,5 miles
Re-arrange this equation: L=length W=width H=height S=surface area S = 2(LW+HW+HL) S/2 = LW+HW+HL (S/2) - LW = HW+HL (S/2) - LW = H(W+L) ((S/2) - LW)/(W+l) = H
Let the side of the square base of the cuboid be s > 0; Let the height of the cuboid be h > 0; Let the surface area be A > 0; Then: A = 2 × (s × s + s × h + h × s) = 2(s² + 2sh) = 2s² + 4sh → 4sh = A - 2s² → h = (A - 2s²)/4s V = s × s × h = s²h = s²(A - 2s²)/4s = s(A - 2s²)/4 = sA/4 - s³/2 This has a maximum value when dV/ds = 0 dV/ds = A/4 - 3s²/2 → 3s²/2 = A/4 → s² = A/6 → s = √(A/6) now: h = (A - 2s²)/4s = s × (A - 2s²)/4s² = s × (A - 2 × A/6)/(4 × A/6) = s × (A - A/3)/(2A/3) = s × (2A/3)/(2A/3) = s × 1 = s → The maximum volume (V) for a cuboid for a given surface area (A) is when the cuboid is a cube.
7.7811/sqrt(2)=7.78If the length of a leg of a 45-45-90 right triangle is s, then the hypotenuse, h, is s*sqrt(2).h=s*sqrt(2) --> s=h/sqrt(2) [This is just a special case of the Pythagorean theory.]
The area is doubled. a,b - cathetus; c - hypotenuse; h - height; S - area. S = (a*b)/2 = (c*h)/2 obviously if k is the doubled height. and A is the new area. A = (c*k)/2 = (c*2h)/2 = c*h and A = S*2