y = 0 gives 0 = - x - 20 so that x = -20 So the intercept is (-20, 0)
(x + 5)(x - 20) x + 5 = 0 x - 20 = 0 x = -5, 20
x2 = x+20 x2-x-20 = 0 (x-5)(x+4) = 0 x = 5 or x = -4
The x intercept is the point where y = 0, the y intercept is the point where x =0: For the x intercept: 5x + 2y = 20, but y = 0 => 5x + 0 = 20 => x = 20/5 = 4. So the x intercept is at (4, 0). For the y intercept: 5x + 2y = 20, but x = 0 [The rest of the calculation is left as an exercise for the reader.]
x2 + 9x + 20 = 0 ∴(x + 4)(x + 5) = 0 ∴x ∈ {-5, -4}
If that's x2 times 20 = 0, then x = 0, because only 0 x 20 = 0
x2 - x = 20 x2 - x - 20 = 0 (x - 5)(x + 4) = 0 x - 5 = 0 and x + 4 = 0 So, x = 5 ans x = -4
y = 0 gives 0 = - x - 20 so that x = -20 So the intercept is (-20, 0)
(x + 5)(x - 20) x + 5 = 0 x - 20 = 0 x = -5, 20
x^2 -x - 20 = 0(x + 4)(x - 5) = 0x + 4 = 0 or x - 5 = 0x = -4 or x = 5
x2 = x+20 x2-x-20 = 0 (x-5)(x+4) = 0 x = 5 or x = -4
0, 20, 40, etc. (just add 20 at a time, or multiply 0 x 20, 1 x 20, 2 x 20, etc.)
The x intercept is the point where y = 0, the y intercept is the point where x =0: For the x intercept: 5x + 2y = 20, but y = 0 => 5x + 0 = 20 => x = 20/5 = 4. So the x intercept is at (4, 0). For the y intercept: 5x + 2y = 20, but x = 0 [The rest of the calculation is left as an exercise for the reader.]
x2 + 9x + 20 = 0 ∴(x + 4)(x + 5) = 0 ∴x ∈ {-5, -4}
x2-35x+300 = 0 (x-15)(x-20) = 0 x = 15 or x = 20
Pseudocode is generally a very loosely defined concept. Various ways you can show your statement: if y = 20 then x = 0 if( y == 20 ) x = 0 if y is 20 then set x to 0
The x intercept is where y = 0, on the x-axis. So plug in y=0 to you equation:-5(x) + 20(0) = 20 ... -5(x) = 20...Divide each side by -5: x = 20/-5 = -20/5 = = -4.Therefore the x intercept is where y = 0 and x = -4 --> (-4, 0)Send any math questions to me for answering!