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konstantin123 [22]
2 years ago
10

point k is between j and l. if jk = x^2 - 4x , kl = 3x - 2 and jl = 28 write and solve an equation to find the lengths of jk and

kl
Mathematics
1 answer:
inn [45]2 years ago
7 0

Answer:

JK=12

KI=16

Step-by-step explanation:

K\in\ [JI]\ \Rightarrow\ |JK|+| KI |=|KI|\\\\x^2-4x+3x-2=28\\\\\Longleftrightarrow\ x^2-x-30=0\\\\\\\Longleftrightarrow\ x^2+5x-6x-30=0\\\\\\\Longleftrightarrow\ x(x+5)-6(x+5)=0\\\\\\\Longleftrightarrow\ (x+5)(x-6)=0\\\\x=-5\ (excluded)\ or\x=6\\\\\\\Longleftrightarrow\ \\|JK|=x^2-4x=6^2-4*6=36-24=12\\|KI|=3x-2=3*6-2=18-2=16\\\\Proof: 12+16=28\\

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13) A rectangular garden has an area of 45 square meters. The garden will be 4 meters longer than its width Write and sowe an eq
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Answer:

9m ×5m

Step-by-step explanation:

Area of a rectangle= length ×width

Let the width be W meters and the length be L meters.

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2 years ago
A cylindrical tank has a base of diameter 12 ft and height 5 ft. The tank is full of water (of density 62.4 lb/ft3).(a) Write do
saw5 [17]

Answer:

a.  71884.8 π lb/ft-s²∫₀⁵(9 - y)dy

b.  23961.6 π lb/ft-s²∫₀⁵(5 - y)dy

c. 99840π lb/ft-s²∫₀⁶rdr

Step-by-step explanation:

.(a) Write down an integral for the work needed to pump all of the water to a point 4 feet above the tank.

The work done, W = ∫mgdy where m = mass of cylindrical tank = ρA([5 + 4] - y) where ρ = density of water = 62.4 lb/ft³, A = area of base of tank = πd²/4 where d = diameter of tank = 12 ft.( we add height of the tank + the height of point above the tank and subtract it from the vertical point above the base of the tank, y to get 5 + 4 - y) and g = acceleration due to gravity = 32 ft/s²

So,

W = ∫mgdy

W = ∫ρA([5 + 4] - y)gdy

W = ∫ρA(9 - y)gdy

W = ρgA∫(9 - y)dy

W = ρgπd²/4∫(9 - y)dy

we integrate W from  y from 0 to 5 which is the height of the tank

W = ρgπd²/4∫₀⁵(9 - y)dy

substituting the values of the other variables into the equation, we have

W = 62.4 lb/ft³π(12 ft)² (32 ft/s²)/4∫₀⁵(9 - y)dy

W = 71884.8 π lb/ft-s²∫₀⁵(9 - y)dy

.(b) Write down an integral for the fluid force on the side of the tank

Since force, F = ∫PdA where P = pressure = ρgh where h = (5 - y) since we are moving from h = 0 to h = 5. So, P = ρg(5 - y)

The differential area on the side of the tank is given by

dA = 2πrdy

So.  F = ∫PdA

F = ∫ρg(5 - y)2πrdy

Since we are integrating from y = 0 to y = 5, we have our integral as

F = ∫ρg2πr(5 - y)dy

F = ∫ρgπd(5 - y)dy    since d = 2r

substituting the values of the other variables into the equation, we have

F = ∫₀⁵62.4 lb/ft³π(12 ft) × 32 ft/s²(5 - y)dy

F = 23961.6 π lb/ft-s²∫₀⁵(5 - y)dy

.(c) How would your answer to part (a) change if the tank was on its side

The work done, W = ∫mgdr where m = mass of cylindrical tank = ρAh where ρ = density of water = 62.4 lb/ft³, A = curved surface area of cylindrical tank = 2πrh  where r = radius of tank, d = diameter of tank = 12 ft. and h =  height of the tank = 5 ft and g = acceleration due to gravity = 32 ft/s²

So,

W = ∫mgdr

W = ∫ρAhgdr

W = ∫ρ(2πrh)hgdr

W = ∫2ρπrh²gdr

W = 2ρπh²g∫rdr

we integrate from r = 0 to r = d/2 where d = diameter of cylindrical tank = 12 ft/2 = 6 ft

So,

W = 2ρπh²g∫₀⁶rdr

substituting the values of the other variables into the equation, we have

W = 2 × 62.4 lb/ft³π(5 ft)² × 32 ft/s²∫₀⁶rdr

W = 99840π lb/ft-s²∫₀⁶rdr

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