Answer:
z<8
Step-by-step explanation:
First: write down the equation : 6z<48
Second: divide both sides with 6 (since its the only number that can be divided for both sides)
Third: then you'll get z<8
Answer:
Step-by-step explanation:
1. Find two numbers that add to make the coefficient of x (in this case, -5) and that multiply to make the constant term multiplied by the coefficient of x^2 (in this case, -2 x 3 = -6)
Two numbers that work are -6 and +1
-6 x +1 = -6
-6 + -1 = -5
2. Split the middle term into the two numbers that you found.
3x^2 -6x +x -2 = 0
I've put the -6 on the left side because in our next step, when we factorise, it will be easier than having the numbers the other way around.
3. Factorise the left side by taking out common factors from each pair. The pairs I'm talking about here are '3x^2 and -6x', and 'x and -2'
3x (x-2) +1 (x-2) = 0
4. You now have two numbers both being multiplied by the term x-2. We can rearrange this equation to give us two brackets being multiplied by each other.
(3x + 1) (x-2) = 0
5. According to the Null Factor Law, if two terms are multiplied together and the result is 0, then one of those terms must be 0. Make both terms equal to 0 and solve each for x.
3x + 1 = 0 x-2 = 0
3x = -1 x = 2
x = -1/3
6. The solutions to this equation are x = 2 and x = -1/3
8 21/100 is your answer! :)
Answer:
221
Step-by-step explanation:
Day 8 is the first day of the second week.
Day 21 is the last day of week 3.
We need to know the n umber of bicycles made from t = 1 to t = 3
The function is b(t) = 110 + 0.5t^2 - 0.9t, where t is in weeks.
We need to integrate the function with the limits of 1 to 3.





Answer: 221