Help please what is this?
1 answer:
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Complete the square for x and y.
7x^2 + 3y^2 - 28x - 12y = -19
7x^2 - 28x + 3y^2 - 12y = -19
7(x^2 - 4x) + 3(y^2 - 4y) = -19
7(x^2 - 4x + 4) + 3(y^2 - 4y + 4) = -19 + 28 + 12
7(x - 2)^2 + 3(y - 2)^2 = 21
7x^2 + 3y^2 - 28x - 12y = -19
7x^2 - 28x + 3y^2 - 12y = -19
7(x^2 - 4x) + 3(y^2 - 4y) = -19
7(x^2 - 4x + 4) + 3(y^2 - 4y + 4) = -19 + 28 + 12
7(x - 2)^2 + 3(y - 2)^2 = 21
Answer:
Step-by-step explanation:
we know that
In this problem ABCD is a rectangle
so
DC=AB
BC=AD
Let
BC ---> the length of rectangle
DC ---> the width of rectangle
we know that
The perimeter of rectangle is equal to
we have
so
simplify
----> equation A
----> equation B
substitute equation B in equation A
Solve for BC
Combine like terms
Divide by 3 both sides
<em>Find the value of DC</em>
---->
Remember that
DC=AB
therefore