For this case, the first thing we must do is define variables.
We have then:
n: number of cans that each student must bring
We know that:
The teacher will bring 5 cans
There are 20 students in the class
At least 105 cans must be brought, but no more than 205 cans
Therefore the inequation of the problem is given by:
Answer:
105 <u><</u> 20n + 5 <u><</u> 205
the possible numbers n of cans that each student should bring in is:
105 <u><</u> 20n + 5 <u><</u> 205
I'll just factor the above equation.
x² + 18x + 80
x² ⇒ x * x
80
can be:
1 x 80
2 x 40
4 x 20
5 x 16
8 x 10 Correct pair
(x+8)(x+10)
x(x+10) +8(x+10) ⇒ x² + 10x + 8x + 80 = x² + 18x + 80
x+8 = 0
x = -8
x+10 = 0
x = -10
x = -8
(-8)² + 18(-8) + 80 = 0
64 - 144 + 80 = 0
144 - 144 = 0
0 = 0
(-10)² + 18(-10) + 80 = 0
100 - 180 + 80 = 0
180 - 180 = 0
0 = 0
I think the algebra tiles will not be a good tool to use to factor the quadratic equation because the equation is not a perfect square quadratic equation.
4:6 x:18
Lets solve for x
Cross multiply
4*18 = 72
6*x = 6x
6x = 72
6x/6 = 72/6
x = 12
Answer: x = 12
Answer:
5. f(x) = 10,000 (1.5)^x
Step-by-step explanation:
We would have to multiply the original amount by 1.50^x because the initial amount would be 1, and 50% increase would be .5 so 1.5 and you raise it to the number of years to show the total increase.
Let's test it.
Initial:
10,000
After 1 year
10,000 + (.5*10000)
10,000 + 5000 = 15,000
After 2 years
15,000 + (.5*15000)
15,000 + 7500 = 22,500
Let's try our equation.
f(x) = 10,000 (1.5)^x
x = 2
10,000(1.5)^2
10,000(2.25) = 22,500
The same!
