Answer:
In the form of
Y= mx+c
Y= 1/2x +2
m = 1/2
Step-by-step explanation:
A linear equation in it's standard form is in the format
Y= mx+c
Where m is the slope and c is the y intercept
Let's use these two points to determine both the slope and the equation
(2, 3), (4,4)
Slope= (y2-y1)/(x2-x1)
Slope= (4-3)/(4-2)
Slope= 1/2
Equation of the linear function
(Y-y1)/(x-x1)= m
(Y-3)/(x-2)= 1/2
2(y-3) = x-2
2y -6 = x-2
2y= x-2+6
2y= x+4
Y= 1/2x +2
Check the picture below.
notice that the car exits the northbound highway, and whilst going at 50mph for 1.5 hours, that simply means going for 75 miles.
make sure your calculator is in Degree mode.
Answer:
The domain is the set of x values for which the function resides in.
In this graph, it is unclear as to if the function goes on to infinity, but if it does, the answer is quite easily:
(-2, 4] and [7, ∞)
since the function starts on the left at -2 then continues to the right, with a pause, then indefinitely after.
Answer: B. There are more boys at Mark's school than at Leslie's school because the ratio 41 to 48 is greater than the ratio 11 to 12.
Step-by-step explanation:
Here are the options:
A There are more boys at Mark's school than at Leslie's school because the ratio 11 to 12 is greater than the ratio 41 to 48.
B. There are more boys at Mark's school than at Leslie's school because the ratio 41 to 48 is greater than the ratio 11 to 12.
C. There are more boys at Leslie's school than at Mark's school because the ratio 41 to 48 is greater than the ratio 11 to 12.
At leslie's school the ratio of boys and girls is 11 to 12. This implies that the fraction of boys in the school to total students will be:
= 11/(11 + 12) = 11/23 = 0.4783
At Marks school the ratio of boys to girls is 41 to 48. Thus implies that the fraction of boys in the school to total students will be:
= 41 / (41 + 48) = 41/85= 0.4824
Based on the calculation, we can deduce that there are more boys at Mark's school than at Leslie's school because the ratio 41 to 48 is greater than the ratio 11 to 12.
First we need to find the gradient of K
which is y1-y2/x1-x2
(-1,3) and (5,-2)
so it becomes 3-(-2)/-1-5
m=-5/6
when two lines are perpendicular their gradients multiply to make -1
that means the gradient of L has to be 6/5
we can substitute the point on L (5,-2) and the gradient of 6/5 into y=mx+c
-2 = (6/5) x 5 + c
c = -8
the equation of line L is y= 6x/5 -8
