Answer:
![0.07](https://tex.z-dn.net/?f=0.07)
Step-by-step explanation:
we know that
The probability that a point chosen randomly inside the rectangle is in the square is equal to divide the area of the square by the area of rectangle
Let
x-----> the area of square
y----> the area of rectangle
P -----> the probability
![P=\frac{x}{y}](https://tex.z-dn.net/?f=P%3D%5Cfrac%7Bx%7D%7By%7D)
<em>Find the area of square (x)</em>
![A=5^{2}=25\ in^{2}](https://tex.z-dn.net/?f=A%3D5%5E%7B2%7D%3D25%5C%20in%5E%7B2%7D)
<em>Find the area of rectangle (y)</em>
![A=25*15=375\ in^{2}](https://tex.z-dn.net/?f=A%3D25%2A15%3D375%5C%20in%5E%7B2%7D)
Find the probability P
![P=\frac{25}{375}=0.07](https://tex.z-dn.net/?f=P%3D%5Cfrac%7B25%7D%7B375%7D%3D0.07)
Answer:
A
Step-by-step explanation:
The area (A) of a triangle is calculated as
A =
bh ( b is the base and h the perpendicular height )
Here b = 13 and h = 4 , thus
A =
× 13 × 4 =
× 52 = 26 units² → A
Let's put it into numbers:
Ruba: 1 000 000
Zamba: 100 000 000 000
Zamba has many more aliens than Ruba!!!
it has 100 000 000 000/1 000 000 =100 000 times more aliens,
in absolute numbers: 100 000 000 000 - 1000 000 = =99 999 900 000
100 000 000 000
- 100 000
=99 999 900 000
Answer:
Step-by-step explanation:
We are to show that ![\sqrt{3} cosec140^{0} - sec140^{0} = 4\\](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D%20cosec140%5E%7B0%7D%20-%20sec140%5E%7B0%7D%20%3D%204%5C%5C)
<u>Proof:</u>
From trigonometry identity;
![cosec \theta = \frac{1}{sin\theta} \\sec\theta = \frac{1}{cos\theta}](https://tex.z-dn.net/?f=cosec%20%5Ctheta%20%3D%20%5Cfrac%7B1%7D%7Bsin%5Ctheta%7D%20%5C%5Csec%5Ctheta%20%3D%20%5Cfrac%7B1%7D%7Bcos%5Ctheta%7D)
![\sqrt{3} cosec140^{0} - sec140^{0} \\= \frac{\sqrt{3} }{sin140} - \frac{1}{cos140} \\= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\](https://tex.z-dn.net/?f=%5Csqrt%7B3%7D%20cosec140%5E%7B0%7D%20-%20sec140%5E%7B0%7D%20%5C%5C%3D%20%5Cfrac%7B%5Csqrt%7B3%7D%20%7D%7Bsin140%7D%20-%20%5Cfrac%7B1%7D%7Bcos140%7D%20%5C%5C%3D%20%5Cfrac%7B%5Csqrt%7B3%7Dcos140-sin140%20%7D%7Bsin140cos140%7D%20%5C%5C)
From trigonometry, 2sinAcosA = Sin2A
![= \frac{\sqrt{3}cos140-sin140 }{sin140cos140} \\\\= \frac{\sqrt{3}cos140-sin140 }{sin280/2}\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{2sin280}\\\\= \frac{4(\sqrt{3}/2cos140-1/2sin140) }{sin280}\\since sin420 = \sqrt{3}/2 \ and \ cos420 = 1/2 \\ then\\\frac{4(sin420cos140-cos420sin140) }{sin280}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%5Csqrt%7B3%7Dcos140-sin140%20%7D%7Bsin140cos140%7D%20%5C%5C%5C%5C%3D%20%20%5Cfrac%7B%5Csqrt%7B3%7Dcos140-sin140%20%7D%7Bsin280%2F2%7D%5C%5C%3D%20%20%5Cfrac%7B4%28%5Csqrt%7B3%7D%2F2cos140-1%2F2sin140%29%20%7D%7B2sin280%7D%5C%5C%5C%5C%3D%20%5Cfrac%7B4%28%5Csqrt%7B3%7D%2F2cos140-1%2F2sin140%29%20%7D%7Bsin280%7D%5C%5Csince%20sin420%20%3D%20%5Csqrt%7B3%7D%2F2%20%5C%20and%20%5C%20cos420%20%3D%201%2F2%20%20%5C%5C%20then%5C%5C%5Cfrac%7B4%28sin420cos140-cos420sin140%29%20%7D%7Bsin280%7D)
Also note that sin(B-C) = sinBcosC - cosBsinC
sin420cos140 - cos420sin140 = sin(420-140)
The resulting equation becomes;
![\frac{4(sin(420-140)) }{sin280}](https://tex.z-dn.net/?f=%5Cfrac%7B4%28sin%28420-140%29%29%20%7D%7Bsin280%7D)
= ![\frac{4sin280}{sin280}\\ = 4 \ Proved!](https://tex.z-dn.net/?f=%5Cfrac%7B4sin280%7D%7Bsin280%7D%5C%5C%20%3D%204%20%5C%20Proved%21)