Answer:
An equation of a line that passes through the point (4,3) and is perpendicular to the graph of the equation will be:
Step-by-step explanation:
We know that the slope-intercept form of the line equation is
![y=mx+b](https://tex.z-dn.net/?f=y%3Dmx%2Bb)
where
Given the line
y = -13x+4
comparing with the slope-intercept form of the line equation
The slope of the line = m = -13
We know that a line perpendicular to another line contains a slope that is the negative reciprocal of the slope of the other line, such as:
slope = m = -13
Thus, the slope of the the new perpendicular line = – 1/m = -1/-13 = 1/13
Using the point-slope form of the line equation
![y-y_1=m\left(x-x_1\right)](https://tex.z-dn.net/?f=y-y_1%3Dm%5Cleft%28x-x_1%5Cright%29)
where
- m is the slope of the line
substituting the values of the slope m = 1/13 and the point (4, 3)
![y-y_1=m\left(x-x_1\right)](https://tex.z-dn.net/?f=y-y_1%3Dm%5Cleft%28x-x_1%5Cright%29)
![y-3=\frac{1}{13}\left(x-4\right)](https://tex.z-dn.net/?f=y-3%3D%5Cfrac%7B1%7D%7B13%7D%5Cleft%28x-4%5Cright%29)
Add 3 to both sides
![y-3+3=\frac{1}{13}\left(x-4\right)+3](https://tex.z-dn.net/?f=y-3%2B3%3D%5Cfrac%7B1%7D%7B13%7D%5Cleft%28x-4%5Cright%29%2B3)
![y=\frac{1}{13}x-\frac{4}{13}+3](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1%7D%7B13%7Dx-%5Cfrac%7B4%7D%7B13%7D%2B3)
![y=\frac{1}{13}x+\frac{35}{13}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B1%7D%7B13%7Dx%2B%5Cfrac%7B35%7D%7B13%7D)
Therefore, an equation of a line that passes through the point (4,3) and is perpendicular to the graph of the equation will be:
Answer:
a=e
Step-by-step explanation:
a+abcdef b=asd d+da
Answer:
153.3852
Step-by-step explanation:
After the decimal, you count the places. 3 is in the ones place, 8 in the tenths, 5 in the hundreths, and 1 in the thousandths. The thousandths place is what we are rounding, so we look at the next number, 9, which tells us to round the 1 up to 2
There's <u>24 possible outcomes</u>.
There's only 6 outcomes with the cube and 4 outcomes with the cards. Multiply 6 and 4 and you get 24.
There's nothing preventing us from computing one integral at a time:
![\displaystyle \int_0^{2-x} xyz \,\mathrm dz = \frac12xyz^2\bigg|_{z=0}^{z=2-x} \\\\ = \frac12xy(2-x)^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E%7B2-x%7D%20xyz%20%5C%2C%5Cmathrm%20dz%20%3D%20%5Cfrac12xyz%5E2%5Cbigg%7C_%7Bz%3D0%7D%5E%7Bz%3D2-x%7D%20%5C%5C%5C%5C%20%3D%20%5Cfrac12xy%282-x%29%5E2)
![\displaystyle \int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy = \frac12\int_0^{1-x}xy(2-x)^2\,\mathrm dy \\\\ = \frac14xy^2(2-x)^2\bigg|_{y=0}^{y=1-x} \\\\= \frac14x(1-x)^2(2-x)^2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_0%5E%7B1-x%7D%5Cint_0%5E%7B2-x%7Dxyz%5C%2C%5Cmathrm%20dz%5C%2C%5Cmathrm%20dy%20%3D%20%5Cfrac12%5Cint_0%5E%7B1-x%7Dxy%282-x%29%5E2%5C%2C%5Cmathrm%20dy%20%5C%5C%5C%5C%20%3D%20%5Cfrac14xy%5E2%282-x%29%5E2%5Cbigg%7C_%7By%3D0%7D%5E%7By%3D1-x%7D%20%5C%5C%5C%5C%3D%20%5Cfrac14x%281-x%29%5E2%282-x%29%5E2)
![\displaystyle\int_0^1\int_0^{1-x}\int_0^{2-x}xyz\,\mathrm dz\,\mathrm dy\,\mathrm dx = \frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_0%5E1%5Cint_0%5E%7B1-x%7D%5Cint_0%5E%7B2-x%7Dxyz%5C%2C%5Cmathrm%20dz%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20dx%20%3D%20%5Cfrac14%5Cint_0%5E1x%281-x%29%5E2%282-x%29%5E2%5C%2C%5Cmathrm%20dx)
Expand the integrand completely:
![x(1-x)^2(2-x)^2 = x^5-6x^4+13x^3-12x^2+4x](https://tex.z-dn.net/?f=x%281-x%29%5E2%282-x%29%5E2%20%3D%20x%5E5-6x%5E4%2B13x%5E3-12x%5E2%2B4x)
Then
![\displaystyle\frac14\int_0^1x(1-x)^2(2-x)^2\,\mathrm dx = \left(\frac16x^6-\frac65x^5+\frac{13}4x^4-4x^3+2x^2\right)\bigg|_{x=0}^{x=1} \\\\ = \boxed{\frac{13}{240}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac14%5Cint_0%5E1x%281-x%29%5E2%282-x%29%5E2%5C%2C%5Cmathrm%20dx%20%3D%20%5Cleft%28%5Cfrac16x%5E6-%5Cfrac65x%5E5%2B%5Cfrac%7B13%7D4x%5E4-4x%5E3%2B2x%5E2%5Cright%29%5Cbigg%7C_%7Bx%3D0%7D%5E%7Bx%3D1%7D%20%5C%5C%5C%5C%20%3D%20%5Cboxed%7B%5Cfrac%7B13%7D%7B240%7D%7D)