To find the total area of this polygon, you would have to find the areas of the rectangle and two triangles and add them up.
Area of Triangle = 1/2 x base x height
= 1/2 x 16 x 6
= 8 x 6
= 48
And since there are two triangles, we can multiply 48 by 2 to get the areas of both triangles.
Area of Both Triangles = 48 x 2 = 96
Area of Rectangle = base x height
= 16 x 10 = 160
Total Area = 96 + 160 = 256 cm2
In the given triangle , Coordinates of A (-1,3) , B (-5,-1), C (3,-1).
Let's find the slope of AB and AC .
Formula of slope is
![m = \frac{y_{2} - y_{1} }{x_{2} - x_{1}}](https://tex.z-dn.net/?f=%20m%20%3D%20%5Cfrac%7By_%7B2%7D%20-%20y_%7B1%7D%20%7D%7Bx_%7B2%7D%20-%20x_%7B1%7D%7D%20)
For AB, slope is,
![m = \frac{-1-3}{-5+1} = \frac{-4}{-4} =1](https://tex.z-dn.net/?f=%20m%20%3D%20%5Cfrac%7B-1-3%7D%7B-5%2B1%7D%20%3D%20%5Cfrac%7B-4%7D%7B-4%7D%20%3D1%20)
And slope of AC is
![\frac{-1-3}{3+1} = \frac{-4}{4} = -1](https://tex.z-dn.net/?f=%20%5Cfrac%7B-1-3%7D%7B3%2B1%7D%20%3D%20%5Cfrac%7B-4%7D%7B4%7D%20%3D%20-1%20)
Product of the slopes of AB and AC is
![= 1* (-1) =-1](https://tex.z-dn.net/?f=%20%3D%201%2A%20%28-1%29%20%3D-1%20)
Since the product of the two slopes is -1, so AB and AC are perpendicular.
And if they are perpendicular, angle BAC is 90 degree, so the triangle is right triangle .
Now we check the length of the sides of the triangle, and for that we use distance formula , which is
![d = \sqrt{ (x_{2} - x_{1} )^2 + (y_{2} - y_{1} )^2}](https://tex.z-dn.net/?f=%20d%20%3D%20%5Csqrt%7B%20%28x_%7B2%7D%20-%20x_%7B1%7D%20%29%5E2%20%2B%20%28y_%7B2%7D%20-%20y_%7B1%7D%20%29%5E2%7D%20)
So for AB,
![AB = \sqrt{ (-3-1)^2 + (-5+1)^2} = \sqrt{32} = 4 \sqrt 2](https://tex.z-dn.net/?f=%20AB%20%3D%20%5Csqrt%7B%20%28-3-1%29%5E2%20%2B%20%28-5%2B1%29%5E2%7D%20%3D%20%5Csqrt%7B32%7D%20%3D%204%20%5Csqrt%202%20)
For AC,
![AC = \sqrt{ (-1-3)^2 + (3+1)^2 } =\sqrt{32} = 4 \sqrt 2](https://tex.z-dn.net/?f=%20AC%20%3D%20%5Csqrt%7B%20%28-1-3%29%5E2%20%2B%20%283%2B1%29%5E2%20%7D%20%3D%5Csqrt%7B32%7D%20%3D%204%20%5Csqrt%202%20)
For BC,
![\sqrt{(-1+1)^2 + (3+5)^2 } = 8](https://tex.z-dn.net/?f=%20%5Csqrt%7B%28-1%2B1%29%5E2%20%2B%20%283%2B5%29%5E2%20%7D%20%3D%208%20)
And since two sides are equal, so the triangle is isosceles triangle .
which of the following is the conjugate of a complex number with 2 as the real part and -8i as the imaginary part
<u>Answer-</u>
<em>The probability that in the box there are 1 red and 8 blue markers is</em><em> 0.111 or 11.1%</em>
<u>Solution-</u>
In the box all markers are red or blue. There are total of 9 markers.
So the number of possible combination number of red or blue marker is,
![S=[(0,9),(1, 8),(2,7),(3,6),(4,5),(5,4),(6,3),(7,2),(8,1),(9,0)}]](https://tex.z-dn.net/?f=S%3D%5B%280%2C9%29%2C%281%2C%208%29%2C%282%2C7%29%2C%283%2C6%29%2C%284%2C5%29%2C%285%2C4%29%2C%286%2C3%29%2C%287%2C2%29%2C%288%2C1%29%2C%289%2C0%29%7D%5D)
As at the random drawn, there was a Blue marker, so the condition of (9,0) i.e 9 Red marker and 0 Blue marker is not a case.
So the sample space becomes,
![S=[(0,9),(1, 8),(2,7),(3,6),(4,5),(5,4),(6,3),(7,2),(8,1)}]](https://tex.z-dn.net/?f=S%3D%5B%280%2C9%29%2C%281%2C%208%29%2C%282%2C7%29%2C%283%2C6%29%2C%284%2C5%29%2C%285%2C4%29%2C%286%2C3%29%2C%287%2C2%29%2C%288%2C1%29%7D%5D)
![|S|=9](https://tex.z-dn.net/?f=%7CS%7C%3D9)
Let us assume that E is the event that in the box there are 1 red and 8 blue markers. So
![E=[(1,8)]](https://tex.z-dn.net/?f=E%3D%5B%281%2C8%29%5D)
![|E|=9](https://tex.z-dn.net/?f=%7CE%7C%3D9)
The probability that in the box there are 1 red and 8 blue markers is,
![P(\text{1 Red and 8 Blue})=\dfrac{|E|}{|S|}=\dfrac{1}{9}=0.111=11.1\%](https://tex.z-dn.net/?f=P%28%5Ctext%7B1%20Red%20and%208%20Blue%7D%29%3D%5Cdfrac%7B%7CE%7C%7D%7B%7CS%7C%7D%3D%5Cdfrac%7B1%7D%7B9%7D%3D0.111%3D11.1%5C%25)
Step-by-step explanation:
amplitude is 3. so the sin will start at zero and go up to 3 and down to -3. asymptote are at 3 and -3
it's period didn't change it's 2π
it's phase shifted horizontally π/3 to the left
it's phase shifted vertically 2 units down