Answer:
{-8+5/7a} to {-4+6/8b}
Step-by-step explanation:
Answer:
4.5 meters/second
Step-by-step explanation:
First we can find out how to convert one hour into seconds, which will be 3600 seconds. Now we have to figure out how many meters are in a mile, and multiply that by 10 to figure out how many meters are traveled per 3600 seconds (hour). The amount of meters in a mile is about 1609.34, which means the number of meters in 10 miles is 16093.4 meters. Now we know that 10 miles per hour is equal to 16093.4 meters per 3600 seconds. The last thing we need to do is divide the amount of meters per 3600 seconds by 3600 to find the amount of meters traveled in one second. This means we have to do 16093.4/3600 which is about 4.47, which rounded to the nearest tenth is 4.5.
Answer:
these numbers are limited and you can make sure using a calculator
The average rate of change is given by [f(b) - f(a)]/(b - a), where a and b represent the interval [a, b].
Let a = -2
Let b = -1
[(-1)^2 + 10 - ((-2)^2 + 10)]/(-1 -(-2))
[1 + 10 - (4 + 10)]/(-1 + 2)
[1 + 10 -4 -10]/(1)
[1 - 4]/(1)
-3/1
Answer: -3
For this case we must indicate which of the equations shown can be solved using the quadratic formula.
By definition, the quadratic formula is applied to equations of the second degree, of the form:
![ax ^ 2+ bx+ c = 0](https://tex.z-dn.net/?f=ax%20%5E%202%2B%20bx%2B%20c%20%3D%200)
Option A:
![2x ^ 2-3x +10 = 2x + 21](https://tex.z-dn.net/?f=2x%20%5E%202-3x%20%2B10%20%3D%202x%20%2B%2021)
Rewriting we have:
![2x ^ 2-3x-2x+ 10-21 = 0\\2x ^ 2-5x-11 = 0](https://tex.z-dn.net/?f=2x%20%5E%202-3x-2x%2B%2010-21%20%3D%200%5C%5C2x%20%5E%202-5x-11%20%3D%200)
This equation can be solved using the quadratic formula
Option B:
![2x ^ 2-6x-7 = 2x ^ 2](https://tex.z-dn.net/?f=2x%20%5E%202-6x-7%20%3D%202x%20%5E%202)
Rewriting we have:
![2x ^ 2-2x ^ 2-6x-7 = 0\\-6x-7 = 0](https://tex.z-dn.net/?f=2x%20%5E%202-2x%20%5E%202-6x-7%20%3D%200%5C%5C-6x-7%20%3D%200)
It can not be solved with the quadratic formula.
Option C:
![5x ^ 2 + 2x-4 = 2x ^ 2](https://tex.z-dn.net/?f=5x%20%5E%202%20%2B%202x-4%20%3D%202x%20%5E%202)
Rewriting we have:
![5x ^ 2-2x ^ 2 + 2x-4 = 0\\3x ^ 2 + 2x-4 = 0](https://tex.z-dn.net/?f=5x%20%5E%202-2x%20%5E%202%20%2B%202x-4%20%3D%200%5C%5C3x%20%5E%202%20%2B%202x-4%20%3D%200)
This equation can be solved using the quadratic formula
Option D:
![5x ^ 3-3x + 10 = 2x ^ 2](https://tex.z-dn.net/?f=5x%20%5E%203-3x%20%2B%2010%20%3D%202x%20%5E%202)
Rewriting we have:
![5x ^ 3-2x ^ 2-3x + 10 = 0](https://tex.z-dn.net/?f=5x%20%5E%203-2x%20%5E%202-3x%20%2B%2010%20%3D%200)
It can not be solved with the quadratic formula.
Answer:
A and C