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Nuetrik [128]
3 years ago
11

2. the student to faculty ratio at a small college is 17:3 the total of students and faculty is 740 how many faculty members are

there at college? how many students?
3. the speedy fast ski resort has started to keep track of the number of skiers and snowboarders who bought season passes. the ratio of the number of skiers who bought season passes to the number of snowboarders who bought season passes is 1:2. if 1259 more snowboarders bought season passer than skiers, how many snowboarders ams how many skiers bought season passes?
Mathematics
1 answer:
balu736 [363]3 years ago
6 0
<span>#2) 629 students and 111 faculty. #3) 1259 skiers and 2518 snowboarders.

Explanation<span>:
#1) Let S be the number of students and F be the number of faculty. S/F = 17/3.
We know that S+F = 740; subtracting F from both sides, we have S=740-F. We can use this to rewrite the proportion:
(740-F)/F=17/3.

Cross multiply:
3(740-F)=17(F).

Using the distributive property, we have
3*740-3*F=17F; 2220-3F=17F.

Add 3F to both sides:
2220-3F+3F=17F+3F
2220=20F.

Divide both sides by 20:
2220/20 = 20F/20
111=F.

There are 111 faculty members. Substitute that into our equation, S=740-F: S=740-111
S=629.

#2) Let x be the number of skiers. We know that there were 1259 more snowboarders with passes than skiers; that gives us x+1259.
The ratio of skiers to snowboarders is 1/2; this gives us the proportion x/(x+1259)=1/2.

Cross multiply:
2(x)=1(x+1259)
2x=x+1259.

Subtract x from both sides
2x-x=x+1259-x
x=1259.

There were 1259 skiers. This means there were 1259+1259=2518 snowboarders.</span></span>
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Please help me out guys. The photo will show you what to do. 50 points cause I need it done fast
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Answer:

a) starting height: 5.5 ft

b) hang time: 5.562 seconds

c) maximum height: 126.5 ft

d) time to maximum height: 2.75 seconds

Step-by-step explanation:

a) The starting height is the height at t=0.

h(0) = -16·0 +88·0 +5.5

h(0) = 5.5

The starting height is 5.5 feet.

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b) The ball is in the air between t=0 and the non-zero time when h(t) = 0. We can find the latter by solving ...

-16t^2 + 8t +5.5 = 0

t^2 -(11/2)t = 5.5/16 . . . . . subtract 5.5, then divide by -16

t^2 -(11/2)t +(11/4)^2 = (5.5/16) +(11/4)^2 . . . . complete the square

(t -11/4)^2 = 126.5/16 . . . . . . . . . . . . . . . . . . . . call this [eq1] for later use

t -11/4 = √7.90625

t = 2.75 +√7.90625 ≈ 5.562

The ball will be in the air about 5.562 seconds.

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c) If we multiply [eq1] above by -16 and add the constant on the right, we get the vertex form of the height equation:

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The vertex at (2.75, 126.5) tells us ...

The maximum height of the ball is 126.5 feet.

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d) That same vertex point tells us ...

The maximum height will be reached at t = 2.75 seconds.

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If you really need answers fast, a graphing calculator can give them to you in very short order (less than a minute).

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