X = gym memberships, and y = bottles of water
30x + 7y = 1050.....when x = 28
30(28) + 7y = 1050
840 + 7y = 1050
7y = 1050 - 840
7y = 210
y = 210/7
y = 30....so he would have to sell 30 water bottles
Answer:
$0 < p ≤ $25
Step-by-step explanation:
We know that coach Rivas can spend up to $750 on 30 swimsuits.
This means that the maximum cost that the coach can afford to pay is $750, then if the cost for the 30 swimsuits is C, we have the inequality:
C ≤ $750
Now, if each swimsuit costs p, then 30 of them costs 30 times p, then the cost of the swimsuits is:
C = 30*p
Then we have the inequality:
30*p ≤ $750.
To find the possible values of p, we just need to isolate p in one side of the inequality.
So we can divide both sides by 30 to get:
(30*p)/30 ≤ $750/30
p ≤ $25
And we also should add the restriction:
$0 < p ≤ $25
Because a swimsuit can not cost 0 dollars or less than that.
Then the inequality that represents the possible values of p is:
$0 < p ≤ $25
$10.50x + $200 = y
X represents the number of hours she works at her regular job
Y represents the total amount of money made over the weekend
Answer:
Step-by-step explanation:
Tn = a + (n-1)d
a = 3 and d = 1.5×3 = 4.5
T2 = a + d = 3 + 4.5 = 7.5
T3 = a + 2d = 3 + 2(4.5) = 12
T4 = a + 3d = 3 + 3(4.5) = 16.5
