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atroni [7]
3 years ago
7

Add​​ using a number line.

Mathematics
1 answer:
coldgirl [10]3 years ago
5 0

Please refer to the attached image.

-2/5, 4/5 and -2/5 + 4/5 have been marked in red in the image

The points marked on the number line are separated by a distance of 1/5.

So when you add 4/5 to -2/5 you need to move 1/5 4 times (4/5 = 4 x 1/5) towards the right to reach the SUM.

Based on this, the sum of -2/5 and 4/5 is 2/5, and has been highlighted in the image.

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5 POINTS !! ANSWER FASTTT
MariettaO [177]
N= 2

Substitute 2 in for n in expression

=(5 + n) ÷ (n - 1)
=(5 + 2) ÷ (2 - 1)
=7 ÷ 1
=7

If n= 2, the expression equals 7.

Hope this helps! :)
3 0
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

7 0
3 years ago
What is the definition for compound interest
Vitek1552 [10]
Compound interest is the addition of interest to the principal sum of a loan or deposit, or in other words, interest on interest.
8 0
3 years ago
What's the first step of solving -x^2-4x+16=0 ??? NOW!
MaRussiya [10]

Answer:

Step-by-step explanation:

Use the Quadratic Formula.

-x² - 4x + 16 = 0

x = [4±√(4²-4(-1)(16)]/[2(-1)] = [4±√80]/(-2) = [4±4√5]/(-2) = -2±2√5

5 0
2 years ago
Let f(x) = √x-5<br><br> Find f^ -1.
iris [78.8K]

Answer:

f^{-1}(x)= x^2 + 5 \,,\,\,x\geq 0

Step-by-step explanation:

Use the method of swapping x and y and isolating y.

f(x) = y = \sqrt{x-5}\,,\,\,\,x\geq 5,y\geq0\\\\f^{-1}:x=\sqrt{y-5}\,,\,\,\,y\geq 5, x\geq 0\\x^2 = y-5\\y = x^2 + 5 = f^{-1}(x)\\

Note that domain/range will be swapped as well.

5 0
3 years ago
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