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:
y * ( 6*5)
Step-by-step explanation:
(y x 6) x 5
We can change the order of multiplication by changing where the parentheses are placed using the associative property
y * ( 6*5)
Answer:
hat: $12
t-shirt: $8
Step-by-step explanation:
Let the price of 1 hat = h.
Let the price of 1 t-shirt = t.
Julie: h + 2t = 28
Raj: 2h + t = 32
We have a system of 2 equations in 2 variables.
h + 2t = 28
2h + t = 32
Let's use the substitution method to solve the system of equations.
Solve the first equation for h.
h = 28 - 2t
Substitute 28 - 2t for h in the second equation.
2(28 - 2t) + t = 32
56 - 4t + t = 32
56 - 3t = 32
-3t = -24
t = 8
h = 28 - 2t
h = 28 - 2(6)
h = 12
Answer:
hat: $12
t-shirt: $8
Answer: 2x-y=15
Step-by-step explanation: can u be more detailin