Step-by-step explanation:
2y - 4x + 6 = 0
2y = 4x - 6
y = 4x/2 - 6/2
y = 2x - 3
Answer: 195
Step-by-step explanation:
(0.3x + 66) + (0.1x + 36) = 180. ( 180 because thats the angle of a straight line)
0.3x + 0.1x + 66 + 36 = 180
0.4x + 102 = 180
0.4x = 180 - 102
0.4x = 78
x = 195
The first integral has a well-known beta function representation, so the second one should too. The beta function itself is defined as
![B(x,y) = \displaystyle \int_0^1 t^{x-1} (1-t)^{y-1} \, dt](https://tex.z-dn.net/?f=B%28x%2Cy%29%20%3D%20%5Cdisplaystyle%20%5Cint_0%5E1%20t%5E%7Bx-1%7D%20%281-t%29%5E%7By-1%7D%20%5C%2C%20dt)
and satisfies the identity
![\displaystyle B(x,y) = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20B%28x%2Cy%29%20%3D%20%5Cfrac%7B%5CGamma%28x%29%20%5CGamma%28y%29%7D%7B%5CGamma%28x%2By%29%7D)
Later on, we'll also use the so-called reflection formula for the gamma function; for non-integer z,
![\Gamma(z) \Gamma(1-z) = \dfrac{\pi}{\sin(\pi z)}](https://tex.z-dn.net/?f=%5CGamma%28z%29%20%5CGamma%281-z%29%20%3D%20%5Cdfrac%7B%5Cpi%7D%7B%5Csin%28%5Cpi%20z%29%7D)
as well as the identity
![\dfrac{\Gamma(z+1)}{\Gamma(z)} = z](https://tex.z-dn.net/?f=%5Cdfrac%7B%5CGamma%28z%2B1%29%7D%7B%5CGamma%28z%29%7D%20%3D%20z)
Replace
in both integrals, so that
![\displaystyle \int_0^{\frac\pi2} \sqrt{\sin(x)} \, dx = \int_0^1 \frac{\sqrt x}{\sqrt{1-x^2}} \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E%7B%5Cfrac%5Cpi2%7D%20%5Csqrt%7B%5Csin%28x%29%7D%20%5C%2C%20dx%20%3D%20%5Cint_0%5E1%20%5Cfrac%7B%5Csqrt%20x%7D%7B%5Csqrt%7B1-x%5E2%7D%7D%20%5C%2C%20dx)
![\displaystyle \int_0^{\frac\pi2} \frac{dx}{\sqrt{\sin(x)}} = \int_0^1 \frac{dx}{\sqrt x \sqrt{1-x^2}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E%7B%5Cfrac%5Cpi2%7D%20%5Cfrac%7Bdx%7D%7B%5Csqrt%7B%5Csin%28x%29%7D%7D%20%3D%20%5Cint_0%5E1%20%5Cfrac%7Bdx%7D%7B%5Csqrt%20x%20%5Csqrt%7B1-x%5E2%7D%7D)
Now replace
:
![\displaystyle \int_0^1 \frac{\sqrt x}{\sqrt{1-x^2}} \, dx = \frac12 \int_0^1 x^{-\frac14} (1-x)^{-\frac12} \, dx = \frac12 B\left(\frac34, \frac12\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E1%20%5Cfrac%7B%5Csqrt%20x%7D%7B%5Csqrt%7B1-x%5E2%7D%7D%20%5C%2C%20dx%20%3D%20%5Cfrac12%20%5Cint_0%5E1%20x%5E%7B-%5Cfrac14%7D%20%281-x%29%5E%7B-%5Cfrac12%7D%20%5C%2C%20dx%20%3D%20%5Cfrac12%20B%5Cleft%28%5Cfrac34%2C%20%5Cfrac12%5Cright%29%20)
![\displaystyle \int_0^1 \frac{dx}{\sqrt x \sqrt{1-x^2}} = \frac12 \int_0^1 x^{-\frac34} (1-x)^{-\frac12} \, dx = \frac12 B\left(\frac14, \frac12\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E1%20%5Cfrac%7Bdx%7D%7B%5Csqrt%20x%20%5Csqrt%7B1-x%5E2%7D%7D%20%3D%20%5Cfrac12%20%5Cint_0%5E1%20x%5E%7B-%5Cfrac34%7D%20%281-x%29%5E%7B-%5Cfrac12%7D%20%5C%2C%20dx%20%3D%20%5Cfrac12%20B%5Cleft%28%5Cfrac14%2C%20%5Cfrac12%5Cright%29)
So, the original integral (which I condense here to a double integral) is
![\displaystyle \int_0^{\frac\pi2} \int_0^{\frac\pi2} \sqrt{\frac{\sin(x)}{\sin(y)}} \, dx \, dy = \frac14 B\left(\frac34, \frac12\right) B\left(\frac14, \frac12\right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E%7B%5Cfrac%5Cpi2%7D%20%5Cint_0%5E%7B%5Cfrac%5Cpi2%7D%20%5Csqrt%7B%5Cfrac%7B%5Csin%28x%29%7D%7B%5Csin%28y%29%7D%7D%20%5C%2C%20dx%20%5C%2C%20dy%20%3D%20%5Cfrac14%20B%5Cleft%28%5Cfrac34%2C%20%5Cfrac12%5Cright%29%20B%5Cleft%28%5Cfrac14%2C%20%5Cfrac12%5Cright%29)
![\displaystyle = \frac14 \frac{\Gamma\left(\frac14\right) \Gamma\left(\frac34\right) \Gamma\left(\frac12\right)^2}{\Gamma\left(\frac54\right) \Gamma\left(\frac34\right)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%3D%20%5Cfrac14%20%5Cfrac%7B%5CGamma%5Cleft%28%5Cfrac14%5Cright%29%20%5CGamma%5Cleft%28%5Cfrac34%5Cright%29%20%5CGamma%5Cleft%28%5Cfrac12%5Cright%29%5E2%7D%7B%5CGamma%5Cleft%28%5Cfrac54%5Cright%29%20%5CGamma%5Cleft%28%5Cfrac34%5Cright%29%7D)
![\displaystyle = \frac14 \frac{\Gamma\left(\frac14\right) \Gamma\left(\frac34\right) \Gamma\left(\frac12\right)^2}{\frac14 \Gamma\left(\frac14\right) \Gamma\left(\frac34\right)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%3D%20%5Cfrac14%20%5Cfrac%7B%5CGamma%5Cleft%28%5Cfrac14%5Cright%29%20%5CGamma%5Cleft%28%5Cfrac34%5Cright%29%20%5CGamma%5Cleft%28%5Cfrac12%5Cright%29%5E2%7D%7B%5Cfrac14%20%5CGamma%5Cleft%28%5Cfrac14%5Cright%29%20%5CGamma%5Cleft%28%5Cfrac34%5Cright%29%7D)
![\displaystyle = \Gamma\left(\frac12\right)^2 = \frac{\pi}{\sin\left(\frac\pi2\right)} = \boxed{\pi}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%3D%20%5CGamma%5Cleft%28%5Cfrac12%5Cright%29%5E2%20%3D%20%5Cfrac%7B%5Cpi%7D%7B%5Csin%5Cleft%28%5Cfrac%5Cpi2%5Cright%29%7D%20%3D%20%5Cboxed%7B%5Cpi%7D)
Answer:
I do not agree. If Mr. Evans worked 42.25 at a rate of $9.20 per hour for the first 40 hours and 1.5 times that rate after, he would have made a total of $399.05.
Step-by-step explanation:
In order to find the total amount that Mr. Evans would have earned last week, we need to first calculate how much he earned in his regular pay at 40 hours. For up to 40 hours, Mr. Evans makes $9.20 per hour, so his total wages earned after 40 hours is: 40 x 9.2 = $368.00. However, Mr. Evans also worked 2.25 hours of overtime at a rate of 1.5 times $9.20 or 9.2 x 1.5 = $13.80. At $13.80 for 2.25 hours, Mr. Evans would earn: 13.80 x 2.25 = $31.05. When you add his base pay of $368.00 plus his over time pay of $31.05, you get a total of $399.05.
m<3 = m<7 true
m1 = m<8 true
the others are not true since c is not parallel