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deff fn [24]
3 years ago
11

I travelled at 60km/h and took 2 hours for a certain journey. How long would it have taken me if I had travelled at 50km/h?​

Mathematics
1 answer:
Marina86 [1]3 years ago
7 0

Answer:

2 hours and 24 minutes

Step-by-step explanation:

2 hours at 60 km/h means you have travelled 2*60=120 km

120 km at 50 km/h takes 120/50 = 2.4 hours

2.4 hours is 2 hours and 0.4*60 = 24 minutes.

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Elenna [48]
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4 0
2 years ago
Max and Sasha exercise a total of 20 hours each week. Max exercises 15 hours less than 4 times the
navik [9.2K]

Answer:

Max exercises 13 hours a week, and Sasha 7.

Step-by-step explanation:

To find the number of hours each of them exercises during the week, we solve the system of equations.

In the second equation:

x = 4y - 15

Replacing in the first equation:

x + y = 20

4y - 15 + y = 20

5y = 35

y = \frac{35}{5}

y = 7

So Sasha exercises 7 hours per week.

Max:

x + y = 20

x + 7 = 20

x = 13

Max exercises 13 hours a week.

6 0
3 years ago
Find mzP<br><br> A 50<br> B. 60<br> C. 70<br> D. 120
marishachu [46]

Answer:

d.120

Step-by-step explanation:

p=70

m=120

z=120

8 0
3 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
Is (1, 3) a solution to the system of inequalities below?
Oxana [17]
<h3>Substitute and Check:</h3>

y > 2x + 1 \\ 3 > 2(1) + 1 \\ 3 > 3 \\ false

<h3>We can stop here and conclude that ( 1 , 3 ) is not a solution to this system, since it does not even satisfy the first equation, but check for the other one just in case:</h3>

y <  - 3x \\ 3 <  - 3(1) \\3 <  - 3 \\ also \: false

8 0
2 years ago
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