how many three a 3 8's are in 1 ?
3x8? 24.
(sqrt((8/8)+8))! = 6 Since: 8 divided by 8 equals 1 1 + 8 = 9; the square root of 9 is 3; 3 factorial equals 6. Alternatively, there is: cube_root(8) + cube_root(8) + cube_ root(8) = 6
8s = 72 divide both sides by 8 s = 9
8s=ps=p/8
Following order of operations, multiplication happens first and addition happens second. -5(-8s-3)+2(5s+4) =40s+15+10s+8 =50s+23 50s+23 cannot be added as they are not like terms. Like terms need to have the same variables.
8
1
There are 0 eight's in three, but there are two and two third threes in eight. Hope this helps :)
There are 50 8s in 400
There are 70 8s in 560
3 of them with a remainder of 4 or 28/8 = 3.5
8 - 8/8 = 8 - 1 = 7
A = (s, 2s), B = (3s, 8s) The midpoint of AB is C = [(s + 3s)/2, (2s + 8s)/2] = [4s/2, 10s/2] = (2s, 5s) Gradient of AB = (8s - 2s)/(3s - s) = 6s/2s = 3 Gradient of perpendicular to AB = -1/(slope AB) = -1/3 Now, line through C = (2s, 5s) with gradient -1/3 is y - 5s = -1/3*(x - 2s) = 1/3*(2s - x) or 3y - 15s = 2s - x or x + 3y = 17s
27 / 8
64 = 8 x 8 !
Points: (s, 2s) and (3s, 8s) Slope: (8s-2s)/(3s-s) = 6s/2s = 3 Perpendicular slope: -1/3 Midpoint: (s+3s)/2 and (2s+8s)/2 = (2s, 5s) Equation: y-5s = -1/3(x-2s) => 3y-15s = -1(x-2s) => 3y = -x+17x Perpendicular bisector equation in its general form: x+3y-17s = 0
It is found as follows:- Points: (s, 2s) and (3s, 8s) Slope: (2s-8s)/(s-3s) = -6s/-2s = 3 Perpendicular slope: -1/3 Midpoint: (s+3s)/2 and (2s+8s)/2 = (2s, 5s) Equation: y-5s = -1/3(x-2s) Multiply all terms by 3: 3y-15s = -1(x-2s) => 3y = -x+17s In its general form: x+3y-17s = 0