Hey there!
The domain and range for G are as follows:
Domain: -9, -9, -8, 4
Range: -6, -4, 0, 4
Hope this answer helps you out!
(5^2-1)/3
(25-1)/3
24/3
8
Your answer is 8
Answer:
Part A)
![\displaystyle \frac{dy}{dx}=-\frac{2xy+y^2}{x^2+2xy}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdx%7D%3D-%5Cfrac%7B2xy%2By%5E2%7D%7Bx%5E2%2B2xy%7D)
Part B)
![\displaystyle y=-\frac{5}{8}x+\frac{9}{4}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D-%5Cfrac%7B5%7D%7B8%7Dx%2B%5Cfrac%7B9%7D%7B4%7D)
Step-by-step explanation:
We have the equation:
![\displaystyle x^2y+y^2x=6](https://tex.z-dn.net/?f=%5Cdisplaystyle%20x%5E2y%2By%5E2x%3D6)
Part A)
We want to find the derivative of our function, dy/dx.
So, we will take the derivative of both sides with respect to <em>x:</em>
<em />
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The derivative of a constant is 0. We can expand the left:
![\displaystyle \frac{d}{dx}\Big[x^2y\Big]+\frac{d}{dx}\Big[y^2x\Big]=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%5CBig%5Bx%5E2y%5CBig%5D%2B%5Cfrac%7Bd%7D%7Bdx%7D%5CBig%5By%5E2x%5CBig%5D%3D0)
Differentiate using the product rule:
![\displaystyle \Big(\frac{d}{dx}\big[x^2\big]y+x^2\frac{d}{dx}\big[y\big]\Big)+\Big(\frac{d}{dx}\big[y^2\big]x+y^2\frac{d}{dx}\big[x\big]\Big)=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5CBig%28%5Cfrac%7Bd%7D%7Bdx%7D%5Cbig%5Bx%5E2%5Cbig%5Dy%2Bx%5E2%5Cfrac%7Bd%7D%7Bdx%7D%5Cbig%5By%5Cbig%5D%5CBig%29%2B%5CBig%28%5Cfrac%7Bd%7D%7Bdx%7D%5Cbig%5By%5E2%5Cbig%5Dx%2By%5E2%5Cfrac%7Bd%7D%7Bdx%7D%5Cbig%5Bx%5Cbig%5D%5CBig%29%3D0)
Implicitly differentiate:
![\displaystyle (2xy+x^2\frac{dy}{dx})+(2y\frac{dy}{dx}x+y^2)=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%282xy%2Bx%5E2%5Cfrac%7Bdy%7D%7Bdx%7D%29%2B%282y%5Cfrac%7Bdy%7D%7Bdx%7Dx%2By%5E2%29%3D0)
Rearrange:
![\displaystyle \Big(x^2\frac{dy}{dx}+2xy\frac{dy}{dx}\Big)+(2xy+y^2)=0](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5CBig%28x%5E2%5Cfrac%7Bdy%7D%7Bdx%7D%2B2xy%5Cfrac%7Bdy%7D%7Bdx%7D%5CBig%29%2B%282xy%2By%5E2%29%3D0)
Isolate the dy/dx:
![\displaystyle \frac{dy}{dx}(x^2+2xy)=-(2xy+y^2)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdx%7D%28x%5E2%2B2xy%29%3D-%282xy%2By%5E2%29)
Hence, our derivative is:
![\displaystyle \frac{dy}{dx}=-\frac{2xy+y^2}{x^2+2xy}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdx%7D%3D-%5Cfrac%7B2xy%2By%5E2%7D%7Bx%5E2%2B2xy%7D)
Part B)
We want to find the equation of the tangent line at (2, 1).
So, let's find the slope of the tangent line using the derivative. Substitute:
![\displaystyle \frac{dy}{dx}_{(2,1)}=-\frac{2(2)(1)+(1)^2}{(2)^2+2(2)(1)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdx%7D_%7B%282%2C1%29%7D%3D-%5Cfrac%7B2%282%29%281%29%2B%281%29%5E2%7D%7B%282%29%5E2%2B2%282%29%281%29%7D)
Evaluate:
![\displaystyle \frac{dy}{dx}_{(2,1)}=-\frac{4+1}{4+4}=-\frac{5}{8}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bdy%7D%7Bdx%7D_%7B%282%2C1%29%7D%3D-%5Cfrac%7B4%2B1%7D%7B4%2B4%7D%3D-%5Cfrac%7B5%7D%7B8%7D)
Then by the point-slope form:
![y-y_1=m(x-x_1)](https://tex.z-dn.net/?f=y-y_1%3Dm%28x-x_1%29)
Yields:
![\displaystyle y-1=-\frac{5}{8}(x-2)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y-1%3D-%5Cfrac%7B5%7D%7B8%7D%28x-2%29)
Distribute:
![\displaystyle y-1=-\frac{5}{8}x+\frac{5}{4}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y-1%3D-%5Cfrac%7B5%7D%7B8%7Dx%2B%5Cfrac%7B5%7D%7B4%7D)
Hence, our equation is:
![\displaystyle y=-\frac{5}{8}x+\frac{9}{4}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20y%3D-%5Cfrac%7B5%7D%7B8%7Dx%2B%5Cfrac%7B9%7D%7B4%7D)
Answer:
The answer is on the attachment.
To solve you must find the area of the triangle plus the area of the rectangle:
Area of Triangle = 1/2 base * height
Area of Rectangle = length * width
The area of the rectangle is:
16 * 8 = 128
The area of the Triangle:
Find the base first, since we know that two sides of a rectangle are equal, the top missing portion must equal 16
12 + triangle base = 16
Triangle base = 4
Height = 6
Area = 3 * 4 = 12
Now add:
128 + 12 = 140feet(squared)
Always bring your units