Answer:
32 units^2
Step-by-step explanation:
This is a trapezoid
The area of a trapezoid is found by
A = 1/2 (b1+b2) h
A = 1/2 ( 4+12) *4
A = 1/2 (16) *4
A =32
Answer:
See explanation
Step-by-step explanation:
Use the definitions:
![\tan \alpha=\dfrac{\sin \alpha}{\cos \alpha}\\ \\\cot \alpha=\dfrac{\cos \alpha}{\sin \alpha}\\ \\\sec \alpha=\dfrac{1}{\cos \alpha}\\ \\\csc \alpha=\dfrac{1}{\sin \alpha}\\ \\](https://tex.z-dn.net/?f=%5Ctan%20%5Calpha%3D%5Cdfrac%7B%5Csin%20%5Calpha%7D%7B%5Ccos%20%5Calpha%7D%5C%5C%20%5C%5C%5Ccot%20%5Calpha%3D%5Cdfrac%7B%5Ccos%20%5Calpha%7D%7B%5Csin%20%5Calpha%7D%5C%5C%20%5C%5C%5Csec%20%5Calpha%3D%5Cdfrac%7B1%7D%7B%5Ccos%20%5Calpha%7D%5C%5C%20%5C%5C%5Ccsc%20%5Calpha%3D%5Cdfrac%7B1%7D%7B%5Csin%20%5Calpha%7D%5C%5C%20%5C%5C)
Now,
![\tan^2\alpha -\cot^2\alpha=\dfrac{\sin^2\alpha}{\cos^2\alpha}-\dfrac{\cos^2\alpha}{\sin^2\alpha}=\dfrac{\sin^4\alpha-\cos ^4\alpha}{\sin^2\alpha\cos ^2\alpha }=\\ \\=\dfrac{(\sin^2\alpha-\cos ^2\alpha)(\sin^2\alpha-\cos ^2\alpha)}{\sin^2\alpha\cos ^2\alpha }=\dfrac{(\sin^2\alpha-\cos ^2\alpha)\cdot 1}{\sin^2\alpha\cos ^2\alpha }](https://tex.z-dn.net/?f=%5Ctan%5E2%5Calpha%20-%5Ccot%5E2%5Calpha%3D%5Cdfrac%7B%5Csin%5E2%5Calpha%7D%7B%5Ccos%5E2%5Calpha%7D-%5Cdfrac%7B%5Ccos%5E2%5Calpha%7D%7B%5Csin%5E2%5Calpha%7D%3D%5Cdfrac%7B%5Csin%5E4%5Calpha-%5Ccos%20%5E4%5Calpha%7D%7B%5Csin%5E2%5Calpha%5Ccos%20%5E2%5Calpha%20%7D%3D%5C%5C%20%5C%5C%3D%5Cdfrac%7B%28%5Csin%5E2%5Calpha-%5Ccos%20%5E2%5Calpha%29%28%5Csin%5E2%5Calpha-%5Ccos%20%5E2%5Calpha%29%7D%7B%5Csin%5E2%5Calpha%5Ccos%20%5E2%5Calpha%20%7D%3D%5Cdfrac%7B%28%5Csin%5E2%5Calpha-%5Ccos%20%5E2%5Calpha%29%5Ccdot%201%7D%7B%5Csin%5E2%5Calpha%5Ccos%20%5E2%5Calpha%20%7D)
and
![\sec^2\alpha(1-\cot^2\alpha)=\dfrac{1}{\cos^2 \alpha}\left(1-\dfrac{\cos^2\alpha}{\sin^2\alpha}\right)=\dfrac{1}{\cos^2 \alpha}\left(\dfrac{\sin^2\alpha-\cos^2\alpha}{\sin^2\alpha}\right)=\\ \\=\dfrac{\sin^2\alpha-\cos ^2\alpha}{\sin^2\alpha\cos ^2\alpha}](https://tex.z-dn.net/?f=%5Csec%5E2%5Calpha%281-%5Ccot%5E2%5Calpha%29%3D%5Cdfrac%7B1%7D%7B%5Ccos%5E2%20%5Calpha%7D%5Cleft%281-%5Cdfrac%7B%5Ccos%5E2%5Calpha%7D%7B%5Csin%5E2%5Calpha%7D%5Cright%29%3D%5Cdfrac%7B1%7D%7B%5Ccos%5E2%20%5Calpha%7D%5Cleft%28%5Cdfrac%7B%5Csin%5E2%5Calpha-%5Ccos%5E2%5Calpha%7D%7B%5Csin%5E2%5Calpha%7D%5Cright%29%3D%5C%5C%20%5C%5C%3D%5Cdfrac%7B%5Csin%5E2%5Calpha-%5Ccos%20%5E2%5Calpha%7D%7B%5Csin%5E2%5Calpha%5Ccos%20%5E2%5Calpha%7D)
As you can see, left and right parts simplify to the same expression, so left and right parts are the same.
Hello there!
![b + 2 = -5](https://tex.z-dn.net/?f=b%20%2B%202%20%3D%20-5)
Explanation:
↓↓↓↓↓↓↓↓↓↓↓
First you had to subtract by 2 from both sides of the equation.
![b+2-2=-5-2](https://tex.z-dn.net/?f=b%2B2-2%3D-5-2)
Simplify it should be the correct answer.
![b=-7](https://tex.z-dn.net/?f=b%3D-7)
Answer⇒⇒⇒b=-7
Hope this helps!
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Have a great day!
-Charlie
1.(5.17204021639)
2.(2.91548)
3.(21.14159)
4.(4.55581)
Five angle of hexagon is 115∘
each sum of all angles of hexagon = 720
∴ let sixth angle of hexagon be x
x+115+115+115+115+115=720
x+575=720
x=720−575
x=145∘