Answer:
$541.67 per month
Step-by-step explanation:
Tuition and other expenses = $8,250 per semester.
There are two semesters in a year
She has 4 years to spend
Total semester=4years*2semesters
=8 semesters
4 years in college which is a total of 8 semesters.
Total Tuition and other expenses = $8,250 * 8
= $66,000
She needs a total of $66,00 to complete her college
Assistance from parents=$15,000
Financial aid(per semester)=$4750
Total financial aid=$38,000
Total assistance=
Assistance from parents+ financial aid
=$15000+$38,000
=$53,000
Total savings=Total amount needed - Total assistance
=$66,000 - $53,000
=$13,000
She needs to save $13,000 in two years
There are 12 months in one year
2 years=2*12=24 months
Monthly savings=Total savings/24 months
=$13,000/24
=$541.666666
To the nearest cent
=$541.67
Answer:
Step-by-step explanation:
Graph given in the attachment represents a linear function f(x).
y = -f(x) represents the reflection of the graph across x-axis.
Therefore, y-coordinates of the points on the graph will get opposite notation after the reflection across x-axis.
Points on the graph of function f(x) are,
(0, -2), (2, 0), (4, 2) and (5, 1)
Therefore, points for the graph y = -f(x) will be,
(0, 2), (2, 0), (4, -2) and (5, -1)
Now we can plot these points to get the graph of y = -f(x)
D : 240 pounds.
The problem gives us the ratio of 60lbs : 1 plants. So, if dee plants four plants, each one has 60 lbs, so she’ll have a total of 240 lbs.
Perhaps the easiest way to find the midpoint between two given points is to average their coordinates: add them up and divide by 2.
A) The midpoint C' of AB is
.. (A +B)/2 = ((0, 0) +(m, n))/2 = ((0 +m)/2, (0 +n)/2) = (m/2, n/2) = C'
The midpoint B' is
.. (A +C)/2 = ((0, 0) +(p, 0))/2 = (p/2, 0) = B'
The midpoint A' is
.. (B +C)/2 = ((m, n) +(p, 0))/2 = ((m+p)/2, n/2) = A'
B) The slope of the line between (x1, y1) and (x2, y2) is given by
.. slope = (y2 -y1)/(x2 -x1)
Using the values for A and A', we have
.. slope = (n/2 -0)/((m+p)/2 -0) = n/(m+p)
C) We know the line goes through A = (0, 0), so we can write the point-slope form of the equation for AA' as
.. y -0 = (n/(m+p))*(x -0)
.. y = n*x/(m+p)
D) To show the point lies on the line, we can substitute its coordinates for x and y and see if we get something that looks true.
.. (x, y) = ((m+p)/3, n/3)
Putting these into our equation, we have
.. n/3 = n*((m+p)/3)/(m+p)
The expression on the right has factors of (m+p) that cancel*, so we end up with
.. n/3 = n/3 . . . . . . . true for any n
_____
* The only constraint is that (m+p) ≠ 0. Since m and p are both in the first quadrant, their sum must be non-zero and this constraint is satisfied.
The purpose of the exercise is to show that all three medians of a triangle intersect in a single point.