Answer:
-5x^4+8x^3-2x+1
Step-by-step explanation:
The area of the part of the plane 3x 2y z = 6 that lies in the first octant is mathematically given as
A=3 √(4) units ^2
<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>
Generally, the equation for is mathematically given as
The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

The partial derivatives of a function are f x and f y.

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:
![&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\](https://tex.z-dn.net/?f=%26%3D%5Cint_%7B0%7D%5E%7B2%7D%20%5Cint_%7B0%7D%5E%7B3-%5Cfrac%7B3%7D%7B2%7D%20x%7D%20%5Csqrt%7B%28-3%29%5E%7B2%7D%2B%28-2%29%5E2%2B1dxdy%7D%20%5C%5C%5C%5C%26%3D%5Cint_%7B0%7D%5E%7B2%7D%20%5Cint_%7B0%7D%5E%7B3-%5Cfrac%7B3%7D%7B2%7D%20x%7D%20%5Csqrt%7B14%7D%20d%20x%20d%20y%20%5C%5C%5C%5C%26%3D%5Csqrt%7B14%7D%20%5Cint_%7B0%7D%5E%7B2%7D%5By%5D_%7B0%7D%5E%7B3-%5Cfrac%7B3%7D%7B2%7D%20x%7D%20d%20x%20d%20y%20%5C%5C%5C%5C%26%3D%5Csqrt%7B14%7D%20%5Cint_%7B0%7D%5E%7B2%7D%5Cleft%5B3-%5Cfrac%7B3%7D%7B2%7D%20x%5Cright%5D%20d%20x%20%5C%5C%5C%5C)
![&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}](https://tex.z-dn.net/?f=%26%3D%5Csqrt%7B14%7D%5Cleft%5B3%20x-%5Cfrac%7B3%7D%7B2%7D%20%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%20x%5E%7B2%7D%5Cright%5D_%7B0%7D%5E%7B2%7D%20%5C%5C%5C%5C%26%3D%5Csqrt%7B14%7D%5Cleft%5B3-%5Cfrac%7B3%7D%7B2%7D%20%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%20x%5E%7B2%7D%5Cright%5D_%7B0%7D%5E%7B2%7D%20%5C%5C%5C%5C%26%3D%5Csqrt%7B14%7D%5Cleft%5B3.2-%5Cfrac%7B3%7D%7B2%7D%20%5Ccdot%20%5Cfrac%7B1%7D%7B2%7D%20%5Ccdot%203%5E%7B2%7D%5Cright%5D%20%5C%5C%5C%5C%26%3D3%20%5Csqrt%7B14%7D%20%5Ctext%20%7B%20units%20%7D%7B%20%7D%5E%7B2%7D)
In conclusion, the area is
A=3 √4 units ^2
Read more about the plane
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Answer: Choice D) 5
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Notice the hole at (1,2). This means x = 1 and y = 2
y = f(x)
y = f(1) after we replace x with 1
y = 2
If we go from f(x) to f(x-4), then we shift the xy axis four units to the left. The curve will stay still though it will move relative to the xy axis. The xy axis moves four units to the left, so it appears that the curve moved 4 units to the right.
Add 4 to the x coordinate:
x = 1 ---> x+4 = 1+4 = 5
The y coordinate stays the same. No vertical shifting is done.
So we go from (1,2) to (5,2)
(1,2) is a hole on y = f(x)
(5,2) is a hole on y = f(x-4)
This is why the answer is x = 5
Answer:
Step-by-step explanation:
slopes and equations:
find the equation thur ( 6,1 ) and (-2,-3)
find the slope m
m = (y2-y1) / (x2-x1 )
m = (-3 - 1) / (-2 -6)
m = -4 / -8
m = 1/2
now use the point-slope formula with our known slope
y-y1 = m(x-x1)
y-1 = 1/2(x-6)
y - 1 = 1/2x -3
y = 1/2x -3 +1
y =
x -2
Find the equation parallel to y = 3x + 6 and thur (0,1)
Parallel means the same slope, the slope is 3 for the equation above.
use the slope-intercept formula again with the point given and the slope 3
y-1 = 3(x -0)
y - 1 = 3x
y = 3x +1
Find the equations perpendicular to 2x + y = 8 and the same y intercept as 4y = x + 3.
put both equations into proper form
y = -2x +6
y =
x + 
perpendicular means reciprocal slope and change the sign, the 2nd equation has an intercept of
, so
y = 1/2x + 
there you go Amanda :)
Answer:
10.77m
Step-by-step explanation:
4^2 +10^2=c^2
16+100=116

=10.77