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)
Answer:
The area of the triangle is 
Step-by-step explanation:
Given:
Coordinates D (0, 0), E (1, 1)
Angle ∠DEF = 60°
△DEF is a Right triangle
To Find:
The area of the triangle
Solution:
The area of the triangle is = 
Here the base is Distance between D and E
calculation the distance using the distance formula, we get
DE = 
DE =
DE = 
DE = 
Base = 
Height is DF
DF =
DF = 
DF = 
Now, the area of the triangle is
= 
=
=
=
-26 is not an inequality. It's just a number. Maybe there was
another part of it that you forgot to copy.
Answer:
Step-by-step explanation:
Find the Equation of a Line Given That You Know a Point on the Line And Its Slope. The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept.
Answer:
Slope and y intercept.
Step-by-step explanation:
This is necassary to write the equation of a line in y=mx+b
m is where the slope goes, and b is the y intercept.
Alternatively, you could also write the equation for a line with a point on the line and the slope.
You would then write it in point slope formula:
y-y1=m(x-x1)
This can later be converted into slope intercept form.