Help me! Need answers to questions ii and iii please. Thanks x
1 answer:
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Answer: 24 unit
Step-by-step explanation:
Here, MNOK is a trapezoid, MN=OK, m∠M=60°, NK ⊥ MN , MK=16cm
Since, a mid segment is the line segment which joins the mid points of the equal sides of the isosceles trapezoid.
Let LO is the mid segment of trapezoid MNOK
Where, Let J is the intersection point of KN and LO.
Therefore, LO= LJ+JO --------(1)
And, KL=LO , MO=ON
Since, In triangle KLJ,
∠LKJ=90° ( given), ∠KLJ=60° ( because, LO║ON and it is given m∠M=60°)
Thus, ∠KJL=30°
Therefore, sin 30°=LK/LJ=8/LJ ( Because ΔMKN is a isoceleus triangle where ∠MKN=∠MNK=30°⇒KM=MN)
⇒LJ=8×2=16
Now, In ΔNOJ,
∠ONJ=∠OJN=30°
Therefore, ON=OJ ( by the property of isosceles triangle)
⇒OJ=8
Thus, By putting these values in equation 1) we get,
LO=16+8=24 cm
G(x) = 3x² - 5x + 7
g(-x) = -(3x² - 5x + 7)
g(-x) = -3x² + 5x - 7
This equation cannot be solved because of a few reasons,
1. This equation didn't show that it equals to 0.
2. Even if it equals to zero, square root of a negative number cannot be solved.
(I will show you what I mean)
-3x² + 5x - 7
is the same as
3x² - 5x + 7
by shifting the equation,
for example,
1 - 3 = -2
shifting other side
2 = -1 + 3
using 3x² - 5x + 7 to solve,
a = 3
b = -5
c = 7
∴This equation cannot be solved.
g(-x) = -(3x² - 5x + 7)
g(-x) = -3x² + 5x - 7
This equation cannot be solved because of a few reasons,
1. This equation didn't show that it equals to 0.
2. Even if it equals to zero, square root of a negative number cannot be solved.
(I will show you what I mean)
-3x² + 5x - 7
is the same as
3x² - 5x + 7
by shifting the equation,
for example,
1 - 3 = -2
shifting other side
2 = -1 + 3
using 3x² - 5x + 7 to solve,
a = 3
b = -5
c = 7
∴This equation cannot be solved.