Answer:
The temperature of the 5th day was 3°C
Step-by-step explanation:
Mean temperature of the first 4 days = 8°C
Note that:
Mean = Sum of temperatures ÷ number of days
∴ 8 = sum of temperature ÷ 4


Therefore the sum of the first 4 days = 32
Let the temperature of the next day (the fifth day) be m
Hence,
sum of the temperatures of first 5 days = 32 + m - - - - (1)
Next, the sum of the first 5 days can be calculated from the given average of the first 5 days as follows:
Mean temperature of the first 5 days = 7

Now, you will notice that equation (1) = equation (2)
∴ 32 + m = 35
m = 35 - 32 = 3
therefore, the temperature of the 5th day was 3°C
Answer:
hi
Step-by-step explanation:
Step-by-step explanation:
You can solve systems of equations using either substitution or elimination. For these problems, I recommend elimination. I'll do the first one as an example.
-3x + 16y = 9
-4x + 8y = 12
Multiply the second equation by -2.
8x − 16y = -24
Add to the first equation (notice the y's cancel out).
(-3x + 16y) + (8x − 16y) = 9 − 24
5x = -15
Solve for x.
x = -3
Now you can plug this into either equation to find y.
-3(-3) + 16y = 9
9 + 16y = 9
y = 0
The solution is (-3, 0).
Answer:
Step-by-step explanation:
Answer:
791.68 cm/s
Step-by-step explanation:
The volume flow rate can be interpreted as the integral of fluid velocity over area
![\dot{V} = \int\limits^6_0 {v(r) 2\pi r} \, dr\\\dot{V} = 2\pi\int\limits^6_0 {(25-r^2)r} \, dr\\\dot{V} = 2\pi\int\limits^6_0 {25r-r^3} \, dr\\\\\dot{V} = 2\pi[12.5r^2 - r^4/4]_0^6\\\dot{V} = 2\pi(12.5*6^2 - 6^4/4 - 12.5*0 - 0)\\\dot{V} = 2\pi*126 = 791.68 cm/s](https://tex.z-dn.net/?f=%5Cdot%7BV%7D%20%3D%20%5Cint%5Climits%5E6_0%20%7Bv%28r%29%202%5Cpi%20r%7D%20%5C%2C%20dr%5C%5C%5Cdot%7BV%7D%20%3D%202%5Cpi%5Cint%5Climits%5E6_0%20%7B%2825-r%5E2%29r%7D%20%5C%2C%20dr%5C%5C%5Cdot%7BV%7D%20%3D%202%5Cpi%5Cint%5Climits%5E6_0%20%7B25r-r%5E3%7D%20%5C%2C%20dr%5C%5C%5C%5C%5Cdot%7BV%7D%20%3D%202%5Cpi%5B12.5r%5E2%20-%20r%5E4%2F4%5D_0%5E6%5C%5C%5Cdot%7BV%7D%20%3D%202%5Cpi%2812.5%2A6%5E2%20-%206%5E4%2F4%20-%2012.5%2A0%20-%200%29%5C%5C%5Cdot%7BV%7D%20%3D%202%5Cpi%2A126%20%3D%20791.68%20cm%2Fs)