82/3 x 12 = 26/3 x 12 = 26 x 4 = 104
3/5 = 12/x Cross multiply: 3*x = 5*12 = 60 Divide both sides by 3: x = 20
You cannot solve this single equation. You can either change the subject so that it gives x = 12/y or xy = 12, which is the equation of a rectangular hyperbola.
What is 3x-6y=12
23
82/3 x 12 = 26/3 x 12 = 26 x 4 = 104
45 30/12 = x/18; 12x = 30 x 18; 12x = 540; x = 540 ÷ 12 = 45
4 x 3
2+n
3/5 = 12/x Cross multiply: 3*x = 5*12 = 60 Divide both sides by 3: x = 20
You cannot solve this single equation. You can either change the subject so that it gives x = 12/y or xy = 12, which is the equation of a rectangular hyperbola.
1B7 twelve x 29 twelve = 1B7 twelve x 9 twelve plus 1B7 twelve x 20 twelve First 1B7 twelve x 9 twelve is sum of 100 twelve x 9 twelve + B0 twelve x 9 twelve + 7 twelve x 9 twelve in base 10 7 x 9 = 63 = and that equals 53 twelve Next B0 twelve x 9 twelve is 11x12 in base 10 + 0 base 10 and that's 132 since 132 x 9 = 1188 in base ten. . 1188 = 8 x 144 + 3 x 12 + 0 x 1 so 1188 = 830 twelve Now 100 twelve times 9 twelve = 900 twelve so we have to add 53 twelve + 830 twelve + 900 twelve the ones column adds to 3, then twelves column is 5+3 =8 and the hundredfortyfours column is 8+9 = 17 base 10 which is 15 twelve so B0 twelve times 9 twelve equals 1583 twelve Now we have to determine 1B7 twelve x 20 twelve 1B7 twelve times 2 twelve is 200 twelve plus (22 base 10) x 10 twelve + 14 base 10 and that is 200 twelve + 1A0 twelve + 12 twelve = 3B2 twelve. we need to put a zero after that so it is 3B20 The final product is equal to 1583 twelve + 3B20 twelve and that is 4 [16]A3 twelve. we must get rid of that base 10 sixteen. Sixteen is 14 twelve so we put the 4 and carry the 1 and result is 54A3 twelve
5
What is 3x-6y=12
23
y is directly proportional to x. When y = 15, x = 3 therefore y = 5x When x = 12, then y = 5 x 12 = 60.
The lines are mutually perpendicular.