A) let x=months let y=cost y=mx+b y= 20x+ 60 Remember 20 is the variable which means in this case every month it cost 20$ and 60 is the constant which means you only pay it once
B) let x= Months let y= cost y=10x + 150
C) platinum gym y= 20x+60 y=20(1)+60 y=20+60 y=80 Superfit y=10x+150 y=10(1)+150 y=10+150 y=160 Therefore Tom will pay less after the first month
D) y=y solve for x y=20x+60 y=10x+150 20x+60=10x+150 -10x. -10x 10x+ 60=150 -60. -60 10x=90 10x/10=90/10 X=9 9 months Pick any equation and sub in x Y=10x+150 Y=10(9)+150 Y=90+150 Y=240 240$ Know that you only have to do one equation to find y but if you want to show both you can Therefore there is a period where Tom and Edward paid the same amount, after nine months they would both pay $240
E) if they plan on going to the gym 9 months and less the best deal gym is platinum gym but if they want to go for more then 9 months then the best deal gym is super fit
Answer: x=4 and y=3
Step-by-step explanation:
We start eliminating the same numbers with the same variable. So we take out 5y on both equation. We then subract 3x to 6x and we do the same to the right side. We end up having 3x=12 and then we divide 12/3=4. We plug 4 on the x values in one of the equation to find out the y value on that equation. You end up with y=3 and then plug that in the same equation on both to see if its right. Finally after that its correct.
Answer:
432 in.^2
Step-by-step explanation:
The side of the suitcase is a rectangle. One length is 24 inches. The diagonal of the rectangle is 30 inches long. The diagonal is a hypotenuse of a right triangle. The length is a leg. We need to find the other leg.
We use the Pythagorean theorem,
a^2 + b^2 = c^2
(24 in.)^2 + b^2 = (30 in.)^2
576 in.^2 + b^2 = 900 in.^2
b^2 = 324 in.^2
b = sqrt(324 in^2)
b = 18 in
area of rectangle = length * width
A = 24 in. * 18 in.
A = 432 in.^2
Answer:
27.875
Step-by-step explanation:
Answer:
<u>Given </u>
A
<u>Find the inverse of f(x):</u>
- x = 3 + 6f⁻¹(x)
- 6f⁻¹(x) = x - 3
- f⁻¹(x) = (x - 3) / 6
B
- f · f⁻¹( ∛5/6) =
- f( f⁻¹( ∛5/6)) =
- f((∛5/6 - 3)/6) =
- 3 + 6((∛5/6 - 3)/6) =
- 3 + ∛5/6 - 3 =
- ∛5/6
C
- f · f⁻¹(x) =
- f(f⁻¹(x)) =
- f((x - 3)/6) =
- 3 + 6(x - 3)/6 =
- 3 + x - 3 =
- x
