To find the intercepts of the equation (7x + 2y - 14z = 56), we determine where the equation intersects the axes.
For the x-intercept, set (y = 0) and (z = 0):
(7x = 56 \Rightarrow x = 8), so the x-intercept is ((8, 0, 0)).
For the y-intercept, set (x = 0) and (z = 0):
(2y = 56 \Rightarrow y = 28), so the y-intercept is ((0, 28, 0)).
For the z-intercept, set (x = 0) and (y = 0):
(-14z = 56 \Rightarrow z = -4), so the z-intercept is ((0, 0, -4)).
Thus, the intercepts are ((8, 0, 0)), ((0, 28, 0)), and ((0, 0, -4)).
y-intercept: 2 x-intercept: 4
x + 2y = 10At the x-intercept, y=0:x = 10At the y-intercept, x=0 ==> 2y = 10y = 5
The x intercept would be 7/2 and y intercept would be 7/2.
This means that 2y - 3 = 2y + 4 which is not possible.
2y+26 = 28 2y = 28-26 2y = 2 y = 1
y-intercept: 2 x-intercept: 4
x + 2y = 10At the x-intercept, y=0:x = 10At the y-intercept, x=0 ==> 2y = 10y = 5
The x intercept would be 7/2 and y intercept would be 7/2.
This means that 2y - 3 = 2y + 4 which is not possible.
No 2y-3 is not equals y plus 2AND how old are u?
(3, 2)
2y+26 = 28 2y = 28-26 2y = 2 y = 1
-2y=x+8 is the answer
2y(x+2)=
Set y = 0 -x = 8 so that x = -8 is the x-intercept. Set x = 0 2y = 8 so that y = 4 is the y-intercept.
Exactly (2y + 4) .
x = 11 - 2y and 14 - 2y. This means that 11 = 14. Is there a typo in the question?