Answer:
3 cars
Step-by-step explanation:
If 12 cars are needed to carry 36 students,
<em>12</em><em>c</em><em>a</em><em>r</em><em>s</em><em> = 36</em>
the number of cars to carry 9 students will be:
xcars = 9
cross the two equations
36x = 12 X 9
36x = 108
x = 108/36
x= 3
Therefore, 3 cars are needed to carry 9 students
OR
12 cars will carry 36 students
so 1 car will carry, (36/12)students
therefore, 1 car will carry 3 students
So, for 9 students,
9students = 9/3 = 3
Answer:
Step-by-step explanation:
The key and the most critical step is to draw the line first. Don't do anything else before that.
(2,-8) (-2,-8)
What you find out is that the y value is always - 8.
So the line has no slope and any x value
y = 0*x + b
y = - 8 is the equation of the line.
Answer:
C. ![\frac{f(b)-f(1)}{b-1}=20](https://tex.z-dn.net/?f=%5Cfrac%7Bf%28b%29-f%281%29%7D%7Bb-1%7D%3D20)
General Formulas and Concepts:
<u>Calculus</u>
- Mean Value Theorem (MVT) - If f is continuous on interval [a, b], then there is a c∈[a, b] such that
![f'(c)=\frac{f(b)-f(a)}{b-a}](https://tex.z-dn.net/?f=f%27%28c%29%3D%5Cfrac%7Bf%28b%29-f%28a%29%7D%7Bb-a%7D)
- MVT is also Average Value
Step-by-step explanation:
<u>Step 1: Define</u>
![f(x)=e^{2x}](https://tex.z-dn.net/?f=f%28x%29%3De%5E%7B2x%7D)
f'(c) = 20
Interval [1, b]
<u>Step 2: Check/Identify</u>
Function [1, b] is continuous.
Derivative [1, b] is continuous.
∴ There exists a c∈[1, b] such that ![f'(c)=\frac{f(b)-f(a)}{b-a}](https://tex.z-dn.net/?f=f%27%28c%29%3D%5Cfrac%7Bf%28b%29-f%28a%29%7D%7Bb-a%7D)
<u>Step 3: Mean Value Theorem</u>
- Substitute:
![20=\frac{ f(b)-f(1)}{b-1}](https://tex.z-dn.net/?f=20%3D%5Cfrac%7B%20f%28b%29-f%281%29%7D%7Bb-1%7D)
- Rewrite:
![\frac{ f(b)-f(1)}{b-1}=20](https://tex.z-dn.net/?f=%5Cfrac%7B%20f%28b%29-f%281%29%7D%7Bb-1%7D%3D20)
And we have our final answer!
Answer:
55 pounds
Step-by-step explanation:
Add all the differences to the initial weight
55.75+2.125-3.25-0.5+0.875=55
Idk maybe ask an adult around or a friend