<span>The diagonals of a parallelogram bisect each other.
So we just equate the two equations BE and DE
2x^2 - x = x^2 + 6
2x^2 - x - x^2 - 6 = 0
x^2 - x - 6 = 0
This is quadratic eqn. So we obtain.
(x - 3) ( x + 2) = 0
Setting each to 0 and solving for x, we have that x = 3 and x = -2
So we have two possible values for BD.
But Since BE = DE then we can simply double each
If x = 3 BD = 2 (2(3)^2 - 3) = 2 ( 18 - 3) = 30 units
If x = - 2 BD = 2((-2)^2 -3 ) = 2 (8 - 3) = 10 units</span>
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The answer is: <u>2(k2−4k)(2c+5)</u>
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Step:
* Consider 2ck2+5k2−8ck−20k. Do the grouping 2ck2+5k2−8ck−20k=(2ck2+5k2) +(−8ck−20k), and factor out k2 in the first and −4k in the second group.
* Factor out the common term 2c+5 by using the distributive property.
* Rewrite the complete factored expression.
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The area of a triangle is calculated by multiplying the base and the height, then dividing that value by 2.
(1/2)bh
Answer:
x=5
y=1
Step-by-step explanation:
The substitution method is considered easier because of its simplification.
2x-3y=7 (1)
x+2y=7 (2)
considering equation (2),
x = 7 - 2y (3)
using the value of "x" in (1)
2 (7-2y) - 3y=7
14 - 4y- 3y = 7
14 - 7y = 7
7y = 14 - 7
7y = 7
y = 1 (4)
substituting this value of "y" in (3)
x = 7- 2(1)
x = 5
Y+4=10(x+3)
Y+4=10x+30
Y=10x+26
