Answer:
y = x + 1
Step-by-step explanation:
The gradient of a line can be defined by the equation:
m (gradient) = (y1 – y2 ) ÷ (x1 – x2) ----> "1" and "2" should be in subscript
For (-7,-6) we use x2 and y2 (because this point can be anywhere along a line):
x2 = -7, y2 = -6
Plug these values into the formula above:
m = (y-(-6)) ÷ (x-(-7))
m = (y+6) ÷ (x+7)
At this stage, the equation can't be solved as there are two unknowns. Therefore, the gradient must be found another way. Two lines are parallel if they have the same gradient - in their y=mx+c equations, m will be equal.
x - y=7 is the line alluded to in the question. Rearranging this equation into the line equation format gives:
y = x-7 ---> The gradient (coefficient of x) is 1.
Therefore, the gradient of the other parallel line must also be 1.
This can be substituted into the previous equation to give:
1 = (y+6)÷(x+7)
x+7 = y+6
x+1 = y
Therefore, the answer is y=x+1
Answer:
The area of sector AOB is 75.36 unit²
Step-by-step explanation:
Consider the provided information.
Circle O with minor arc AB=60 degrees and OA =12.
We need to find the area of sector.
Area of the sector: 
Substitute 
and r =12 in above formula.
Hence, the area of sector AOB is 75.36 unit²
Observe the given data distribution table carefully.
The 5th class interval is given as,
The upper limit (UL) and lower limit (LL) of this interval are,
Thus, the upper-class limit of this 5th class is 17.4.
For a linear function, the instantaneous rate of change is everywhere equal to the slope. Thus the rate of change of the function h(x)=2x on the interval 2≤x≤4
The rate of change of the function given will equal to its slope, thus;
slope,m=(y-1-y)/(x_1-x)
=(2*4-2*2)/(4-2)
=(8-4)/2
=4/2
=2
the answer is 2
