C(x) = 200 - 7x + 0.345x^2
Domain is the set of x-values (i.e. units produced) that are feasible. This is all the positive integer values + 0, in case that you only consider that can produce whole units.
Range is the set of possible results for c(x), i.e. possible costs.
You can derive this from the fact that c(x) is a parabole and you can draw it, for which you can find the vertex of the parabola, the roots, the y-intercept, the shape (it open upwards given that the cofficient of x^2 is positive). Also limit the costs to be positive.
You can substitute some values for x to help you, for example:
x y
0 200
1 200 -7 +0.345 = 193.345
2 200 - 14 + .345 (4) = 187.38
3 200 - 21 + .345(9) = 182.105
4 200 - 28 + .345(16) = 177.52
5 200 - 35 + 0.345(25) = 173.625
6 200 - 42 + 0.345(36) = 170.42
10 200 - 70 + 0.345(100) =164.5
11 200 - 77 + 0.345(121) = 164.745
The functions does not have real roots, then the costs never decrease to 0.
The function starts at c(x) = 200, decreases until the vertex, (x =10, c=164.5) and starts to increase.
Then the range goes to 164.5 to infinity, limited to the solutcion for x = positive integers.
We have
tan 12.5 = 60 / adj rearrange as
adj = 60 / tan 12.5 = about 270.64 m
Yes. x=3 and y =9 are the solutions.
Because if you change x=3 and y =9 ,you will have
3+2* 9 =21
3+18 =21
21 =21 (true)
Answer:
Jackie sold 12 cars.
Step-by-step explanation:
If we call the number of cars Oscar sold O, and the number of cars Jackie sold J, we can say the following:
O = J + 6
As Oscar sold 6 cars more than Jackie.
Together, they sold 30 cars.
O + J = 30
Since we know that:
O = J + 6
... we can put this into our previous equation.
O + J = 30
(J + 6) + J = 30
J + J + 6 = 30
2 * J + 6 = 30
Subtract 6 from both sides:
2 * J = 24
Divide both sides by 2:
J = 24 / 2
J = 12
Jackie sold 12 cars.
Let the total sum of the scores of the first class of 35 students be
a. The mean is 74.3
.
So
Also, let the total sum of the scores of the second class of 28 students be
b. The mean is 67.6 .
so
The combined group has 35+28=63 students. The sum of their scores is
a+b=2600.5+1892.8=4493.3
Thus, the mean of the combined group is
