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)]
Learn more here:brainly.com/question/12486387
Answer:
- Q11 - 12 sides
- Q12 - 10 sides
Step-by-step explanation:
<u>Use the formula for sum of interior angles of a regular polygon:</u>
<u>Each angle A measures:</u>
- S/n = 180(n - 2)/n
- A = 180(n - 2)n
- An = 180n - 360
- (180 - A)n = 360
- n = 360/(180 - A)
<h3>Question 11</h3>
<u>Each angle is 150, then finding n:</u>
- n = 360/(180 - 150)
- n = 360/30
- n = 12
<h3>Question 12</h3>
<u>Each angle is 144, find n:</u>
- n = 360/(180 - 144)
- n = 360/36
- n = 10
Answer:switch mikes Luke’s and Alex in between each other and let winter keep her own
Step-by-step explanation:
