Suppose x and y are any two co-prime integers. Then 105*x and 105*y will have a GCF of 105.
First, let's write it out: 2y+23=y+75 1. Subtract y from both sides to get: y+23=75 2. Subtract 23 from both sides to get: y=52 And that's your final answer for y!
3 and 2 over 7 = ( 3*7 + 2 )/ 7 = 23/7 7 and 1 over 15 = ( 7*15 + 1 ) / 15 = 106/15 The least common multiple of 7 and 15 is 105 So, 23/7 + 106/15 = 23*15 / 105 + 106*7 / 105 = 345 / 105 + 742 / 105 = 1087 / 105 = (1050 + 37 ) / 105 = 10 and 37 over 105
Yes. If you were to plot y=23 on a graph, you'd have a straight horizontal line where y=23 (because no matter what the value of x is on the graph, y is always 23). As the line that is plotted is straight, the equation is considered linear.
There are an infinite number of different equations that are satisfied by that point. A few of them are: y = 15/23 x y = x - 8 y = 2x - 31 y = 10x - 215 y = 15 cos(x - 23) in radians y = 15 sin(x + 67) in degrees
100 x 23/105 = 21.9%
Suppose x and y are any two co-prime integers. Then 105*x and 105*y will have a GCF of 105.
105 degrees
equations according to the conditions given are 13*x=y; y-105=105-x; solving for x,we have x=15. verifying: 13*15=195. 195-105=90 105-15=90.
23
If you mean: y = 12x-23 then the slope is 12 and the y intercept is -23
First, let's write it out: 2y+23=y+75 1. Subtract y from both sides to get: y+23=75 2. Subtract 23 from both sides to get: y=52 And that's your final answer for y!
BandB - Bella y bestia - 2008 1-105 is rated/received certificates of: Argentina:Atp
3 and 2 over 7 = ( 3*7 + 2 )/ 7 = 23/7 7 and 1 over 15 = ( 7*15 + 1 ) / 15 = 106/15 The least common multiple of 7 and 15 is 105 So, 23/7 + 106/15 = 23*15 / 105 + 106*7 / 105 = 345 / 105 + 742 / 105 = 1087 / 105 = (1050 + 37 ) / 105 = 10 and 37 over 105
The least common multiple of the numbers 25 and 105 is 525.
-15 = y/7(x both sides by 7)-105 = y
If y - 8 = 15, y = 23