Answer:
The gradient of the straight line that passes through (2, 6) and (6, 12) is 
.
Step-by-step explanation:
Mathematically speaking, lines are represented by following first-order polynomials of the form:
(1)
Where:
- Independent variable.
- Dependent variable.
- Slope.
- Intercept.
The gradient of the function is represented by the first derivative of the function:
Then, we conclude that the gradient of the staight line is the slope. According to Euclidean Geometry, a line can be form after knowing the locations of two distinct points on plane. By definition of secant line, we calculate the slope:
(2)
Where:
, 
- Coordinates of point A.
, 
- Coordinates of point B.
If we know that 
and 
, then the gradient of the straight line is:
The gradient of the straight line that passes through (2, 6) and (6, 12) is .
L=W-2LW=35 substitute in W-2 for L(W-2)(W)=35W2 -2w -35=0(W-5)(W-7)=0therefore w=7 and L=5
1/4 mile per minute is the average speed.
Answer:
x=2/5, y=-24/5. (2/5, -24/5).
Step-by-step explanation:
y=-2x-4
y=1/2x-5
----------------
-2x-4=1/2x-5
-2x-1/2x-4=-5
-4/2x-1/2x=-5+4
-5/2x=-1
5/2x=1
x=1/(5/2)
x=2/5
y=1/2(2/5)-5
y=1/5-5
y=1/5-25/5
y=-24/5
Answer:
The equation –3y = 15 – 4x rewritten in slope-intercept form is
✔ y = (4/3)x – 5
.
The y-intercept is
✔ –5
and the slope of the line is
✔ 4/3
.
Line
✔ B
is the graph of the line –3y = 15 – 4x.
