The expression factorized completely is (h +2k)[(h+2k) + (2k-h)]
From the question,
We are to factorize the expression (h+2k)²+4k²-h² completely
The expression can be factorized as shown below
(h+2k)²+4k²-h² becomes
(h+2k)² + 2²k²-h²
(h+2k)² + (2k)²-h²
Using difference of two squares
The expression (2k)²-h² = (2k+h)(2k-h)
Then,
(h+2k)² + (2k)²-h² becomes
(h+2k)² + (2k +h)(2k-h)
This can be written as
(h+2k)² + (h +2k)(2k-h)
Now,
Factorizing, we get
(h +2k)[(h+2k) + (2k-h)]
Hence, the expression factorized completely is (h +2k)[(h+2k) + (2k-h)]
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Answer:
The inverse of the function is 
.
Step-by-step explanation:
The function provided is:
Let 
.
Then the value of <em>x</em> is:
For the inverse of the function, 
.
⇒
Compute the value of ![f[f^{-1}(x)]](https://tex.z-dn.net/?f=f%5Bf%5E%7B-1%7D%28x%29%5D)
as follows:
![f[f^{-1}(x)]=f[\frac{x-5}{3}]](https://tex.z-dn.net/?f=f%5Bf%5E%7B-1%7D%28x%29%5D%3Df%5B%5Cfrac%7Bx-5%7D%7B3%7D%5D)
![=3[\frac{x-5}{3}]+5\\\\=x-5+5\\\\=x](https://tex.z-dn.net/?f=%3D3%5B%5Cfrac%7Bx-5%7D%7B3%7D%5D%2B5%5C%5C%5C%5C%3Dx-5%2B5%5C%5C%5C%5C%3Dx)
Hence proved that ![f[f^{-1}(x)]=x](https://tex.z-dn.net/?f=f%5Bf%5E%7B-1%7D%28x%29%5D%3Dx)
.
Compute the value of ![f^{-1}[f(x)]](https://tex.z-dn.net/?f=f%5E%7B-1%7D%5Bf%28x%29%5D)
as follows:
![f^{-1}[f(x)]=f^{-1}[3x+5]](https://tex.z-dn.net/?f=f%5E%7B-1%7D%5Bf%28x%29%5D%3Df%5E%7B-1%7D%5B3x%2B5%5D)
Hence proved that ![f^{-1}[f(x)]=x](https://tex.z-dn.net/?f=f%5E%7B-1%7D%5Bf%28x%29%5D%3Dx)
.
To solve for the width simply multiply the two numbers:
width of sidewalk = 12 * (3 7/8)
Where 3 7/8 = 31/8
so calculating,
width of sidewalk = 12 * (31 / 8)
<span>width of sidewalk = 46.5 inches</span>
She 56/92 6-5=1 and 567-1=566
To the nearest penny, c = $6,252.50
Total of payments = amount financed + c = $26,752.50
Total of payments ÷ number of payments = monthly payment = $445.88
