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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:

Step-by-step explanation:
Given
Co ordinates of vertices(1,-6) and (-8,-6)
When two points is given then Length of two points is given by



Perimeter of rectangle is =26 units
Let the other side be x
thus

x+9=13
x=4 units
therefore to get the other two co ordinates
such that the length of that side is 4 units is


Horizontal distance will remain same only vertical distance will change in given co ordinates to obtain the remaining two co ordinates
To verify the above two distance between two points must be 13 units
Remember property of exponents

add exponents of same base
ok so
50,000 is 5 times 10^4
so
50000 times 10^15=
5 times 10^4 times 10^15=
5 times

=
5 times 10^19
Answer:

Step-by-step explanation:
In any right triangle, the Pythagorean theorem states:
, where
and
are two legs of the triangle and
is the hypotenuse.
Plugging in given values, we have:
