If dh = 3x -3 and fh = x+7, find the value of x for which defg must be a parallelogram.
2 answers:
Answer:
Option c is correct
x = 5
Step-by-step explanation:
Properties of parallelogram:
- Opposites sides are congruent.
- Opposites angle are congruent
- Diagonal of parallelogram bisect each other.
As per the statement:
In the given DEFG parallelogram,
DH = 3x-3 units
FH=x+7 units
by properties of parallelogram we have;
DH = FH
Subtract x from both sides we have;
Add 3 to both sides we have;
Divide both sides by 2 we have;
x = 5
Therefore, the value of x for which DEFG must be a parallelogram is, 5
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Answer:
Step-by-step explanation:
Given:
Required:
LCM of the polynomials
SOLUTION:
Step 1: Factorise each polynomial
Step 2: find the product of each factor that is common in both polynomials.
We have the following,
The common factors would be: =>
(this is common in both polynomials, so we would take just one of them as a factor.
and,
Their product =
This is a polynomial of degree 2, and remember that complex roots always come in pairs; therefore, in a second degree polynomial you can only have 0 real roots and 2 complex roots, or 2 real roots and 0 complex roots. There is no way you can have 1 real root and 1 complex root because, as stated before, complex (non-real) roots always come in pairs; so the only way of having 1 real root is dumping the polynomial of degree 2 (
) by changing the 14 to 0. That way we'll left with 
which only has one real root:
