Start by using trig to find the length of the line LJ
The triangle KJL (big right angled triangle) has been given the following dimensions
Hypotenuse =
![8 \sqrt{2}](https://tex.z-dn.net/?f=8%20%5Csqrt%7B2%7D%20)
The adjacent angle is 30 degrees
Since we have the hypotenuse and the angle we must use the equation
opposite = Sin(angle) x Hypotenuse
Opposite= sin30 x
![8 \sqrt{2}](https://tex.z-dn.net/?f=8%20%5Csqrt%7B2%7D%20)
Opposite=
![4 \sqrt{2}](https://tex.z-dn.net/?f=4%20%5Csqrt%7B2%7D%20)
Therefore line LJ is
![4 \sqrt{2}](https://tex.z-dn.net/?f=4%20%5Csqrt%7B2%7D%20)
Now look at the smaller right angled triangle (LMJ)
Hypotenuse is the line LJ which is
![4 \sqrt{2}](https://tex.z-dn.net/?f=4%20%5Csqrt%7B2%7D%20)
The adjacent angle is 45
Since we have hypotenuse and angle we must use the equation opposite = sin(angle) * h
therefore
x=
![4 \sqrt{2}](https://tex.z-dn.net/?f=4%20%5Csqrt%7B2%7D%20)
* sin45= 4
Answer:
2 students from each of the 15 classes
Step-by-step explanation:
Just took the test on Edge
Answer:
1) 5f
2) 45gm-7g
Step-by-step explanation:
hope this helped!
Answer:
(-6, -2)
Step-by-step explanation:
![f(x) = {x}^{2} + 12x + 34](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%7Bx%7D%5E%7B2%7D%20%20%2B%2012x%20%2B%2034)
![= f(x) = {1x}^{2} + 12x + 34](https://tex.z-dn.net/?f=%20%3D%20f%28x%29%20%3D%20%20%7B1x%7D%5E%7B2%7D%20%20%2B%2012x%20%2B%2034)
= a = 1, b = 12
![= x = - \frac{12}{2 \times 1}](https://tex.z-dn.net/?f=%20%3D%20x%20%3D%20%20-%20%20%5Cfrac%7B12%7D%7B2%20%5Ctimes%201%7D%20)
![= x = - \frac{12}{2}](https://tex.z-dn.net/?f=%20%3D%20x%20%3D%20%20-%20%20%5Cfrac%7B12%7D%7B2%7D%20)
= x = -6
![= f(x) = {x}^{2} + 12x + 34.. = - 6](https://tex.z-dn.net/?f=%20%3D%20f%28x%29%20%3D%20%20%7Bx%7D%5E%7B2%7D%20%20%2B%2012x%20%2B%2034..%20%3D%20%20-%206)
![= f( - 6) = - 2](https://tex.z-dn.net/?f=%20%3D%20f%28%20-%206%29%20%3D%20%20-%202)
= (-6, -2)