Answer: ![\begin{bmatrix}\mathrm{Solution:}\:&\:x\le \:-\frac{2}{3}\quad \mathrm{or}\quad \:x\ge \:0\:\\ \:\mathrm{Decimal:}&\:x\le \:-0.66666\dots \quad \mathrm{or}\quad \:x\ge \:0\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:-\frac{2}{3}]\cup \:[0,\:\infty \:)\end{bmatrix}](https://tex.z-dn.net/?f=%5Cbegin%7Bbmatrix%7D%5Cmathrm%7BSolution%3A%7D%5C%3A%26%5C%3Ax%5Cle%20%5C%3A-%5Cfrac%7B2%7D%7B3%7D%5Cquad%20%5Cmathrm%7Bor%7D%5Cquad%20%5C%3Ax%5Cge%20%5C%3A0%5C%3A%5C%5C%20%5C%3A%5Cmathrm%7BDecimal%3A%7D%26%5C%3Ax%5Cle%20%5C%3A-0.66666%5Cdots%20%5Cquad%20%5Cmathrm%7Bor%7D%5Cquad%20%5C%3Ax%5Cge%20%5C%3A0%5C%5C%20%5C%3A%5Cmathrm%7BInterval%5C%3ANotation%3A%7D%26%5C%3A%28-%5Cinfty%20%5C%3A%2C%5C%3A-%5Cfrac%7B2%7D%7B3%7D%5D%5Ccup%20%5C%3A%5B0%2C%5C%3A%5Cinfty%20%5C%3A%29%5Cend%7Bbmatrix%7D)
Step-by-step explanation:
Answer:
5 hours
Step-by-step explanation:
Lillian is deciding between two parking garages.
Let the time required to park be represented by t
A = Amount
From Garage A
A = the amount Garage A would charge if Lillian parks for t hours
B = the amount Garage B would charge if Lillian parks for t hours.
Garage A
Garage A charges an initial fee of $4 to park plus $3 per hour.
A = $4 + $3 × t
A = 4 + 3t
Garage B charges an initial fee of $9 to park plus $2 per hour.
B = $9 + $2 × t
B = 9 + 2t
The hours parked, t, that would make the cost of each garage the same is calculated by equating A to B
A = B
4 + 3t = 9 + 2t
Collect like terms
3t - 2t = 9 - 4
t = 5 hours
Therefore, the hours parked, t, that would make the cost of each garage the same is 5 hours
Answer:
answer is 1=2=3244515/15/152/51//51/1=12900 is the answer all you need to do is get 1700k - 10000 and yes its negative.
Step-by-step explanation:
