Check the picture below.
that's done just with the provided links, and I simply ran a line, so it'll be thereabouts, roughly between 13 and 14, depending on how you hang the line.
anyhow, if you have a TI(texas instruments) 89+ or higher, you can just provide the value pairs of the data and run a quadratic regression, or an exponential.
there are some online regression calculators as well.
L(x) is the width of 8x2- 4x which means the answer is A
Ok first of all we know it's linear and will look like y=mx +c
In this example y=C and X=S
The answer is:
C= (225)S + 5500
So for 12 tons of sugar you substitute S with 12..
C=225(12) + 5500
Answer:
- b (x, x - 3, 0)
- d Infinite Solutions
Step-by-step explanation:
1. A graphing calculator or any of several solvers available on the internet can tell you the reduced row-echelon form of the augmented matrix ...
![\left[\begin{array}{ccc|c}2&-2&-1&6\\-1&1&3&-3\\3&-3&2&9\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D2%26-2%26-1%266%5C%5C-1%261%263%26-3%5C%5C3%26-3%262%269%5Cend%7Barray%7D%5Cright%5D)
is the matrix ...
![\left[\begin{array}{ccc|c}1&-1&0&3\\0&0&1&0\\0&0&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%26-1%260%263%5C%5C0%260%261%260%5C%5C0%260%260%260%5Cend%7Barray%7D%5Cright%5D)
The first row can be interpreted as the equation ...
x -y = 3
x -3 = y . . . . . add y-3
The second row can be interpreted as the equation ...
z = 0
Then the solution set is ...
(x, y, z) = (x, x -3, 0) . . . . matches selection B
__
2. The second equation is 2 times the first equation, so the system of equations is dependent. There are infinite solutions.
Answer:
Step-by-step explanation:
The
average rate of change
of f(x) over an interval between 2 points (a ,f(a)) and (b ,f(b)) is the slope of the
secant line
connecting the 2 points.
To calculate the average rate of change between the 2 points use.
∣
∣
∣
∣
∣
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
a
a
f
(
b
)
−
f
(
a
)
b
−
a
a
a
∣
∣
∣
−−−−−−−−−−−−−−−
f
(
9
)
=
9
2
−
6
(
9
)
+
8
=
35
and
f
(
4
)
=
4
2
−
6
(
4
)
+
8
=
0
The average rate of change between (4 ,0) and (9 ,35) is
35
−
0
9
−
4
=
35
5
=
7
This means that the average of all the slopes of lines tangent to the graph of f(x) between (4 ,0) and (9 ,35) is 7.
