An experiment is carried out 400 times.
1 answer:
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Refer to the diagram below.
Because ray NP bisects ∠MNQ, therefore
∠MNP = ∠PNQ = 2x + 1.
Therefore
∠MNQ = 2*∠PNQ = 2(2x + 1) = 4x + 2.
Because ∠MNQ is given as x² - 10, therefore
x² - 10 = 4x + 2
x² - 4x - 12 = 0
(x + 2 )(x - 6) = 0
x = -2, or x = 6
When x = -2,
∠MNQ = 4*(-2) + 2 = -6°
This answer is not acceptablle, therefore x = -2 should be rejected.
When x = 6,
∠MNQ = 4*6 + 2 = 26°
Answer: x = 6, and ∠MNQ = 26°
Because ray NP bisects ∠MNQ, therefore
∠MNP = ∠PNQ = 2x + 1.
Therefore
∠MNQ = 2*∠PNQ = 2(2x + 1) = 4x + 2.
Because ∠MNQ is given as x² - 10, therefore
x² - 10 = 4x + 2
x² - 4x - 12 = 0
(x + 2 )(x - 6) = 0
x = -2, or x = 6
When x = -2,
∠MNQ = 4*(-2) + 2 = -6°
This answer is not acceptablle, therefore x = -2 should be rejected.
When x = 6,
∠MNQ = 4*6 + 2 = 26°
Answer: x = 6, and ∠MNQ = 26°
We are given : Distance of the swing = 100 feet.
Distance of slide = 80 feet.
Angle between swing and slide = 30 degrees.
We need to find the distance between the swing and the slide.
Distance of swing, distance of slide and distance between the swing and the slide form a triangle.
We can apply cosine law to find the distance between the swing and the slide.
c^2 = a^2 +b^2 - 2ab cos C
c^2 = 100^2 +80^2 - 2(100)(80) cos 30°
c^2 = 10000 + 6400 -2* 8000
c^2 = 16400 - 8000
c^2 = 16400 - 13856
c^2 = 2544
c= 50.44
c = 50 feet approximately.
<h3>Therefore, the approximate distance between the swing and the slide is 50 feet.</h3>If a square-shaped pool with side lengths are 3a^5 then find the area of the pool
solution:
A = (3a^5) x (3a^5)
= 9a^5+5
= 9a^10
The area of the square-shaped pool are 9a^10 based on the given side length which is 3a^5
