5 of them with a remainder of 1
4=4(3s) 4=12s s= 1/3
-6r - s
It is 3.833... (3s repeating).
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
4s-5-3s =s-5 The s and 5 can not subtract because they aren't like terms. Like terms have the same variable.
s = 15
5 of them with a remainder of 1
A pentagon has 5 sides.A regular pentagon has every side the same length.Perimeter = 5 x 3s= 15 s(whatever unit a s is, other than a second.)
Let F = father's age Let S = son's age F = 3S F - 5 = 4(S-5) F - 5 = 4S -20 F = 4S - 15 F = 3S 3S = 4S-15 S = 15, age of son
5 + 2s + s = 3s - s + 8 Combining like terms on the same side: 5 + 3s = 2s + 8 Subtracting 2s from both sides: 5 + s = 8 Subtracting 5 from both sides: s = 3
135=3s +15 120=3s 40=s
The LCM of 3s and s^2 is 3s^2
There are no instances of the digit 3 in the number 16. The number 16 consists of the digits 1 and 6. If we were to break down the number 16 into its individual digits, we would have 1 and 6, with no 3 present.
2s + s + 12 =132 ie 3s = 132 -12 3s = 120 s = 40
4=4(3s) 4=12s s= 1/3
5s - 3 = 3s - 92s = -6s = -3