5 is closets to zero
It's exactly 15 units away from zero. If you consider 15 to be close, then it's close.
That depends how you define "close to". 12/15 = 4/5 = 0.8 So 12/15 is closer to 1 than to 0 or 0.5, but closer to 0.75 than 1.
If you are asking if x=0, y=5 is a solution to 5x-3y=15, then no. 0, -5 would be as if you sub in 0 for x you get 5(0) -3y=15, i.e. -3y=15, i.e. y=-5
4w2 = 17w + 15 so 4w2 - 17w - 15 = 0 4w2 - 20w + 3w - 15 = 0 4w*(w - 5) + 3*(w - 5) = 0 (w - 5)*(4w + 3) = 0 So w - 5 = 0 or 4w + 3 = 0 therefore w = 5 or w = -0.75
x2 + 8x = -15 x2 + 8x + 15 = 0 since 15 = 5 x 3 and 5 + 3 = 8, then (x + 5)(x + 3) = 0 x + 5 = 0 or x + 3 = 0 x = -5 or x = -3
-2
2x2 + 7x - 15 = 02x2 + 10x - 3x - 15 = 0(2x2 - 3x) + (10x - 15) = 0x(2x - 3) + 5(2x - 3) = 0(2x - 3)(x + 5) = 0(2x - 3) = 0 or (x + 5) = 0x = 3/2 or x = -5
To be divisible by 5, the last digit must be 0 or 5.The last digit of 15 is 5, so it is divisible by 5.
2x2-x-15 = 0 When factorised: (2x+5)(x-3) = 0
Points: (0, 5) and (10, -15) Slope: -2
If: (2x+10)(3x-15) = 0 => x^2-25 = 0 => (x-5)(x+5) = 0 So: x = 5 or x = -5
p2 = 2p + 15 p2 - 2p - 15 = 0 (p - 5)*(p + 3) = 0 so p - 5 = 0 or p + 3 = 0 so p = 5 or p = -3