Answer:
Step-by-step explanation:
Gym A has a $150 joining fee and costs $35 per month.
Assuming that Casey wants to attend for x months, the cost of using gym A will be
150 + 35 times x months. It becomes
150 + 35x
Gym B has no joining fee and costs $60 per month.
Again, assuming that Casey wants to attend for x months, the cost of using gym B will be
60 × x months = 60x
A) To determine the number of months that it will both gym memberships to be the same, we will equate them.
150 + 35x = 60x
60x - 35x = 150
25x = 150
x = 150/25 = 6
It will take 6 months for both gym memberships to be the same.
B) If Casey plans to only go to the gym for 5 months,
Plan A will cost 150 + 35×5 = $325
Plan B will cost 60 × 5 = $300
Plan B will be cheaper
Your question does not say what were your options, therefore I will answer generically: in order to understand if a point (ordered pair) is contained in a line, you need to substitute the x-component of the pair in the equation of the line and see if the calculations give you the y-component of the pair.
Example:
Your line is <span> y = 4/3x + 1/3
Let's see if <span>(0, 0) and (2, 3) </span>belong to this line
y</span> = <span>4/3·0 + 1/3 = 1/3 </span>≠ 0
Therefore, the line does not contain (0, 0)
y = 4/3·2 + 1/3 = 9/3 = 3
Therefore, the line contains (2, 3)
Answer:
Honestly um I started a few days ago yet dont know
The only way 3 digits can have product 24 is
1 x 3 x 8 = 241 x 4 x 6 = 242 x 2 x 6 = 242 x 3 x 4 = 24
So the digits comprises of 1,3,8 or 1,4,6, or 2,2,6, or 2,3,4
To be divisible by 3 the sum of the digits must be divisible by 3.
1+ 3+ 8=12, 1+ 4+ 6= 11, 2 +2 + 6=10, 2 +3 + 4=9Of those sums of digits, only 12 and 9 are divisible by 3.
So we have ruled out all but integers whose digits consist of1,3,8, and 2,3,4.
Meanwhile they must be odd they either must end in 1 or 3.
The only ones which can end in 1 are 381 and 831.
The others must end in 3.
They must be greater than 152 which is 225. So the
First digit cannot be 1. So the only way its digits can contain of1,3,8 and close in 3 is to be 813.
The rest must contain of the digits 2,3,4, and the only way they can end in 3 is to be 243 or 423.
So there are precisely five such three-digit integers: 381, 831, 813, 243, and 423.
Answer:
4) C
5) C
Step-by-step explanation:
