Can someone please help me out with this geometry problem? I've been stuck on it for almost an hour.
1 answer:
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Answer:
The value of x = 11 and y = 2
Step-by-step explanation:
Given : parallelogram LMNO, MP = 21 m, LP = (y + 3) m, NP = (3y – 1) m, and OP = (2x – 1) m.
We have to find values of x and y.
Let P be the point of intersection of diagonals OM and LN.
In a parallelogram diagonal bisects at right angles and point of intersection divide diagonal in equal parts.
Thus, OP = MP and LP = PN
OP = MP , substitute the values, we get,
(2x-1) = 21
⇒ 2x = 22
⇒ x = 11
LP = PN , substitute the values, we get,
y + 3 = 3y -1
⇒ 3y - y = 4
⇒ 2y = 4
⇒ y = 2
Thus, the value of x = 11 and y = 2
Answer:
a)
And replacing we got:
And simplifying we got:
b) For this case since the coeficient for the higher degree is 2 then the polynomial is of second degree.
c) We need to remember that the closed property for polynomials tell to us that if we apply any operation between two polynomials we need to obtain and other polynomial. For this special case the property is the sum and after multiply we have another polynomial with a higher degree and then the closed property is satisfied.
Step-by-step explanation:
We know a rectangle has sides measuring (4x + 5) units and (3x + 10) units
Part a
For this case we can find the area like this:
And replacing we got:
And simplifying we got:
Part b
For this case since the coeficient for the higher degree is 2 then the polynomial is of second degree.
Part c
We need to remember that the closed property for polynomials tell to us that if we apply any operation between two polynomials we need to obtain and other polynomial. For this special case the property is the sum and after multiply we have another polynomial with a higher degree and then the closed property is satisfied.