If this were to be graphed, the independent variable would be the price of the ticket for the rides. The dependent variable would be the total cost.
The fair admission is not a variable because it is a constant price for every single person who goes into the fair.
The problem asks to use y to represent the total cost and x to represent the number of ride tickets. In order to fully write out the equation, we have to figure out what the fair admission costs.
43.75 = 1.25(25) + b
*b represents the fair admission
Multiply 1.25 by 25
43.75 = 31.25 + b
Subtract 31.25 to find what b costs.
12.50 = b
The fair admission costs $12.50.
Solution: y = 1.25x + 12.50
Formula for the perimeter of a quarter circle: C = ((pi x 2r) / 4) + 2r
C = ((3.14 x 18) / 4) + 18
C = (56.52 / 4) + 18
C = 14.13 + 18
C = 32.13 miles
Hope this helps! :)
Answer:
7, 7.1, √51, 7.2
Step-by-step explanation:
√51 is 7.14
Answer:
The three numbers are 341, 342, and 343
Step-by-step explanation:
We start by assigning X to the first integer. Since they are consecutive, it means that the 2nd number will be X + 1 and the 3rd number will be X + 2 and they should all add up to 1026. Therefore, you can write the equation as follows:
(X) + (X + 1) + (X + 2) = 1026
To solve for X, you first add the integers together and the X variables together. Then you subtract three from each side, followed by dividing by 3 on each side. Here is the work to show our math:
X + X + 1 + X + 2 = 1026
3X + 3 = 1026
3X + 3 - 3 = 1026 - 3
3X = 1023
3X/3 = 1023/3
X = 341
Which means that the first number is 341, the second number is 341 + 1 and the third number is 341 + 2. Therefore, three consecutive integers that add up to 1026 are 341, 342, and 343.
341 + 342 + 343 = 1026
We know our answer is correct because 341 + 342 + 343 equals 1026 as displayed above.
The distance between the Earth's surface and the upper edge of the Earth's atmosphere would be Q - V.
The distance from the Sun to the Earth is Q. The distance from the Sun to the upper edge of the atmosphere is V. If you subtract V from Q, the remaining distance is that of the Earth's atmosphere. Q - V = the atmosphere
