Answer:
Step-by-step explanation:
This question is incomplete; here is the complete question.
A closed cylindrical can of fixed volume V has radius r. (a) Find the surface area, S, as a function of r. (b) What happens to the value of S approaches to infinity? (c) Sketch a graph of S against r, if V=10 cm³.
A closed cylindrical can of volume V is having radius r and height h.
a). Surface area of a cylinder is given by
S = 2(Area of the circular sides) + Lateral area of the can
S = 2πr² + 2πrh
S = 2πr(r + h)
b). Since surface area is directly proportional to radius of the can
S ∝ r
Therefore, when r approaches to infinity (r → ∞)
c). If V = 10 cm³ Then we have to graph S against r.
From the formula V = πr²h
10 = πr²h
h = 
By placing the value of h in the formula of surface area,
S = 
Now we can get the points to plot the graph,
r -2 -1 0 1 2
S -13.72 -13.72 0 26.28 35.13
Answer: 30 $15 tickets and 20 $35 tickets
Step-by-step explanation:
.X%2BY=50
2.15X%2B35Y=1150
From eq. 1,
15X%2B15Y=750
Subtract from eq. 2,
15X%2B35Y-15X-15Y=1150-750
20Y=400
Y=20
Then,
X%2B20=50
X=30
30 $15 tickets and 20 $35 ticket
Hope it helps
Answer:
-50
Well to get the answer of the product of a number and 9 is -450. You must divide 9 by -450 using long divison.
Hello,
I note (a,b,c) the result of a quarters, b dimes and c pennies:
2 solutions:
106=( 3, 3, 1)=( 1, 8, 1)
106=( 0, 0, 106) but : 100= 0*25+ 0*10+ 100
106=( 0, 1, 96) but : 100= 0*25+ 1*10+ 90
106=( 0, 2, 86) but : 100= 0*25+ 2*10+ 80
106=( 0, 3, 76) but : 100= 0*25+ 3*10+ 70
106=( 0, 4, 66) but : 100= 0*25+ 4*10+ 60
106=( 0, 5, 56) but : 100= 0*25+ 5*10+ 50
106=( 0, 6, 46) but : 100= 0*25+ 6*10+ 40
106=( 0, 7, 36) but : 100= 0*25+ 7*10+ 30
106=( 0, 8, 26) but : 100= 0*25+ 8*10+ 20
106=( 0, 9, 16) but : 100= 0*25+ 9*10+ 10
106=( 0, 10, 6) but : 100= 0*25+ 10*10+ 0
106=( 1, 0, 81) but : 100= 1*25+ 0*10+ 75
106=( 1, 1, 71) but : 100= 1*25+ 1*10+ 65
106=( 1, 2, 61) but : 100= 1*25+ 2*10+ 55
106=( 1, 3, 51) but : 100= 1*25+ 3*10+ 45
106=( 1, 4, 41) but : 100= 1*25+ 4*10+ 35
106=( 1, 5, 31) but : 100= 1*25+ 5*10+ 25
106=( 1, 6, 21) but : 100= 1*25+ 6*10+ 15
106=( 1, 7, 11) but : 100= 1*25+ 7*10+ 5
106=( 1, 8, 1) is good
106=( 2, 0, 56) but : 100= 2*25+ 0*10+ 50
106=( 2, 1, 46) but : 100= 2*25+ 1*10+ 40
106=( 2, 2, 36) but : 100= 2*25+ 2*10+ 30
106=( 2, 3, 26) but : 100= 2*25+ 3*10+ 20
106=( 2, 4, 16) but : 100= 2*25+ 4*10+ 10
106=( 2, 5, 6) but : 100= 2*25+ 5*10+ 0
106=( 3, 0, 31) but : 100= 3*25+ 0*10+ 25
106=( 3, 1, 21) but : 100= 3*25+ 1*10+ 15
106=( 3, 2, 11) but : 100= 3*25+ 2*10+ 5
106=( 3, 3, 1) is good
106=( 4, 0, 6) but : 100= 4*25+ 0*10+ 0
