Answer:
Step-by-step explanation:
A possible word problem for 3x+10 ≤75 could be
Mat has $75 dollars saved in his wallet and wants to use it for go carts. Knowing that the entrance to the park is $10, and each time around the go cart course is $3 what is the maximum times he can go around?
3x+10 ≤ 75, subtract 10 from both sides
3x ≤ 75-10, divide both sides by 3
x ≤ 65/3
or x ≤ 21.(6)
So it can go around 21 times the most.
We can solve this problem using discriminant.
x^2-4x-12's discriminant is
(-4)^2-4*-12=16+48 which is clearly larger than 0
This means that it crosses over the axes 2 times.
In case you don't know what discriminant is, its in equation ax^2+bx+c
the discriminant is b^2-4ac.
If its positive it has 2 crosses with x axis, if negative then 0 crosses, if 0 then 1 cross.
Hope this helped at least a little bit :D
Answer:
$481.54
Step-by-step explanation:
672.33-(102.25+(2)44.27)
672.33-(102.25+88.54)
672.33-190.79
481.54
I believe it’s 1/2 hope it helps
