X-x-2/6=x-7/3+2/3
-2/6+7/3-2/3=x
x= -2/6+5/3
x= 8/6
x= 4/3
a) You are told the function is quadratic, so you can write cost (c) in terms of speed (s) as
... c = k·s² + m·s + n
Filling in the given values gives three equations in k, m, and n.

Subtracting each equation from the one after gives

Subtracting the first of these equations from the second gives

Using the next previous equation, we can find m.

Then from the first equation
[tex]28=100\cdot 0.01+10\cdot (-1)+n\\\\n=37[tex]
There are a variety of other ways the equation can be found or the system of equations solved. Any way you do it, you should end with
... c = 0.01s² - s + 37
b) At 150 kph, the cost is predicted to be
... c = 0.01·150² -150 +37 = 112 . . . cents/km
c) The graph shows you need to maintain speed between 40 and 60 kph to keep cost at or below 13 cents/km.
d) The graph has a minimum at 12 cents per km. This model predicts it is not possible to spend only 10 cents per km.
Answer:
The answer to your questions are in bold
Step-by-step explanation:
a)
C = ![\left[\begin{array}{ccc}-6&6\\-2&4\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-6%266%5C%5C-2%264%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= -24 + 12
= -12
b) -1 7 -4 -1 -1 - 4 7 - 1 - 5 6
-2 -6 -8 8 = -2 - 8 -6 + 8 -10 2
2 -3 2 -7 2 + 2 -3 - 7 4 -10
-1 10 -6 5 -1 - 6 10 + 5 -7 15
Answer:
23.96g
Step-by-step explanation:
Make sure to thank me!!!
Answer:
It would take 19 hours and 36 minutes until there are 1040 bacteria present.
Step-by-step explanation:
Given that in a lab experiment, 610 bacteria are placed in a petri dish, and the conditions are such that the number of bacteria is able to double every 23 hours, to determine how long would it be, to the nearest tenth of an hour, until there are 1040 bacteria present, the following calculation must be performed:
610X = 1040
X = 1040/610
X = 1.7049
2 = 23
1.7049 = X
1.7049 x 23/2 = X
39.2131 / 2 = X
19.6 = X
100 = 60
60 = X
60 x 60/100 = X
36 = X
Therefore, it would take 19 hours and 36 minutes until there are 1040 bacteria present.