(A) We let y = the cost and are told x = the number of people. Since you pay $20 per person, the cost is 20x. That is, y=20x
(B) Again, let the cost =y and the number of people is given as x. You pay $10 per person or 10x plus an additional $50 for the room. That is, y=10x+50
(c) Link to graphs: https://www.desmos.com/calculator but if that doesn't work see the attachment for a screen shot. You just have to put the equations (type them) at left and the graph comes automatically.
(D) The admission price is the same when the two equations are equal. You can find this by setting them equal to each other as such: 20x = 10x+50 and solving for x. However, since you just graphed them the point of intersection (where the lines share/have the same point) gives the information. Remembers that (x,y) = (people, cost). The graphs intersect at (5, 100) so for 5 people the cost is the same and the cost is $100.
(E) For the regular rate we let x = 6 and solve for y (the cost). We get y = 20x which is y = (20)(6)=120. It costs $120 using the regular rate to take 6 people. Now let's use the equation for the group rate again with x = 6. Here we get y = 10x +50 or y = 10(6)+50 = $110. The group rate costs $110.
(F) The cost is the same at 5 people but if there are more than five the group rate is better as we saw in part E. So the regular rate is better for less than 5 people.
(G) Here y = $150. Let us use the group rate formula and solve for x (the number of people). 10x+50 = 150 so 10x = 100 and x = 10. Since 10 is more than 5 this is the better deal. However if you don't believe it or want to double check we can solve for x using y = 150 and the regular rate equation. We get: 20x = 150 so x = 7.5 Since we can't bring half a person we would only be able to bring 7 and that is less than 10 so this is not the best choice. Use the group rate and bring 10 people!
2x - 15
15 is decreased that is the key word here which indicates subtraction.
twice a number with given variable x is 2x
subtraction always goes behind so 2x - 15
9514 1404 393
Answer:
382 square units
Step-by-step explanation:
The central four rectangles down the middle of the net are 9 units wide, and alternate between 8 and 7 units high. Then the area of those four rectangles is ...
9(8+7+8+7) = 270 . . . square units
The rectangles making up the two left and right "wings" of the net are 8 units high and 7 units wide, so have a total area of ...
2×(8)(7) = 112 . . . square units
Then the area of the figure computed from the net is ...
270 +112 = 382 . . . square units
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<em>Additional comment</em>
You can reject the first two answer choices immediately, because they are odd. Each face will have an area that is the product of integers, so will be an integer. There are two faces of each size, so <em>the total area of this figure must be an even number</em>.
You may recognize that the dimensions are 8, 8+1, 8-1. Then the area is roughly that of a cube with dimensions of 8: 6×8² = 384. If you use these values (8, 8+1, 8-1) in the area formula, you find the area is actually 384-2 = 382. That area formula is A = 2(LW +H(L+W)).
Slope intercept form is y=5/3x+5
The slope is 5/3
Max will be missing $1.78 in his account. Is that one of the fee choices?