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kherson [118]
3 years ago
9

A rectangle and a triangle have the same area.

Mathematics
1 answer:
Ivanshal [37]3 years ago
7 0
Area of a rectangle is length times width. Area of a triangle is one-half base times height. Now just plug in everything we know and set the equations equal to each other.

(6)(8) = 0.5(8)(h)

Solve for height

48 = 4h

h = 12 m




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4m+8n-4p+8m+16n+8p=12m+24n+4p
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A rectangular classroom has an area of 300 square feet. The length of the classroom is 5 feet longer than the width, w. Create a
weqwewe [10]

Answer:

w^2 + 5w = 300

Step-by-step explanation:

A= length x width

A =300

Width = w

length = w + 5

Therefore

300 = (w) x (w + 5)

300 = w^2 + 5w

w^2 + 5w = 300

6 0
2 years ago
Solve 2x2 + 5x = 12
Delvig [45]

Answer:

A

Step-by-step explanation:

2x² + 5x = 12

2x² + 5x - 12 = 0

x² + 5x - 24 = 0

factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24. the only pair that adds to 5 is 8 and 3

2x² - 3x + 8x - 12 = 0

split the equation and factor

2x² - 3x = 0 ||| 8x - 12 = 0

x(2x - 3) ||| 4(2x - 3)

therefore the pair is (2x - 3)(x + 4)

2x - 3 = 0

2x = 3

x = 3/2

or

x + 4 = 0

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3 0
2 years ago
Read 2 more answers
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

7 0
3 years ago
How many permutations can be made from the letters m, n, o, p, and q taken 3 at a time? 6 30 60
aleksley [76]
There are a total of 60 permutations makeable from those letters 3 at a time.
8 0
3 years ago
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