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:
Yes, Mina is correct
Step-by-step explanation:
Let the triangles be A and B
Given
Triangle A:
Triangle B:
Required
Is Mina's claim correct?
First, we calculate the third angle in both triangles.
For A:
For B:
For triangle A, the angles are: 34, 57 and 89
For triangle B, the angles are: 34, 57 and 89
<em>Since both triangles have the same angles, then by the postulate of AAA (Angle-Angle-Angle), the triangles are similar.</em>
Answer:
=51.63 m^2
Step-by-step explanation:
First find the area of the triangle
A = 1/2 bh = 1/2(3) * 5 = 7.5
Then find the area of the rectangle
A = l*w = 5*6 = 30
Then find the area of the semi circle, which is 1/2 the area of the circle
The diameter is 6 so the radius is 3
A = 1/2 pi r^2 = 1/2 (3.14) (3) ^2 = 14.13
Add the areas together for the total area
7.5+30+14.13
=51.63
