128 POINTS IF YOU HELP
2 answers:
(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!
(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!
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Answer:
length and width=4
height=8
Step-by-step explanation:
Hello to solve this problem we must propose a system of equations of 3x3, that is to say 3 variables and 3 equations.
Ecuation 1
Leght=Width
.L=W
Ecuation 2
To raise the second equation we consider that the length and width of 4 inches less than the height of the box
H-4=W
Ecuation 3
To establish equation number 3, we find the volume of a prism that is the result of multiplying length, width, and height
LxWxH=128
From ecuation 1(w=h)
solving for H
<em />
<em>Using ecuation 2</em>
H-4=W
Now we find the roots of the equation, 2 of them are imaginary, and only one results in 4
W=4in
L=4in
to find the height we use the ecuation 2
H-4=W
H=4+W
H=4+4=8
H=8IN