Answer:
yes another flavor should be in the bag. because the problem stated that there are three different flavor, and only two flavors were mentioned in the problem. the probability of getting the third flavor is 5/7 i guess, because it think the probability of the second flavor was the answer when you added both the numerator and the denominator of 1/3 by 2. like 1/3+2=3/5, so if you added 2 to both the numerator and the denominator of 3/5, you will get 5/7. but i am not so sure about this, i'm sorry in advance.
Ok so first we find the equation that equals one variable.
2y = -x + 9
3x - 6y = -15
We solve for y.
2y = -x + 9
y = -x/2 + 9/2
Then we plug in this y value into the other equation to keep only one variable so we can solve for it.
3x - 6y = -15
3(-x + 9/2) - 6y = -15
-3x + 27/2 - 6y = -15
-9y + 27/2 = -15
-9y = 3/2
-y = 3/18
y = -3/18
Then we plug in this numerical y-value into the first equation which we found out by solving an equation for y.
y = -x/2 + 9/2
-3/18 = -x/2 + 9/2
-84/18 = -x/2
-x = 9 1/3
x = -28/3
Your answer would be (-28/3, -3/18)
Hope this helps!
Answer:
77. 
Proved
78. ![\sec^{4}x \tan^{2} x = \sec^{2}x [\tan^{2}x + \tan^{4}x ]](https://tex.z-dn.net/?f=%5Csec%5E%7B4%7Dx%20%5Ctan%5E%7B2%7D%20x%20%3D%20%5Csec%5E%7B2%7Dx%20%5B%5Ctan%5E%7B2%7Dx%20%2B%20%5Ctan%5E%7B4%7Dx%20%5D)
Proved
79. ![\cos^{3} x\sin^{2} x = [\sin^{2}x - \sin^{4}x] \cos x](https://tex.z-dn.net/?f=%5Ccos%5E%7B3%7D%20x%5Csin%5E%7B2%7D%20x%20%3D%20%5B%5Csin%5E%7B2%7Dx%20-%20%5Csin%5E%7B4%7Dx%5D%20%5Ccos%20x)
Proved.
80. 
Proved.
Step-by-step explanation:
77. Left hand side
= 
= 
= ![\cot^{4}x [\csc^{2}x - 1]](https://tex.z-dn.net/?f=%5Ccot%5E%7B4%7Dx%20%5B%5Ccsc%5E%7B2%7Dx%20-%201%5D)
{Since we know, 
}
= 
= Right hand side (Proved)
78. Left hand side
= 
= ![\sec^{2} x [1 + \tan^{2}x] \tan^{2} x](https://tex.z-dn.net/?f=%5Csec%5E%7B2%7D%20x%20%5B1%20%2B%20%5Ctan%5E%7B2%7Dx%5D%20%5Ctan%5E%7B2%7D%20x)
{Since 
}
= ![\sec^{2}x [\tan^{2}x + \tan^{4}x ]](https://tex.z-dn.net/?f=%5Csec%5E%7B2%7Dx%20%5B%5Ctan%5E%7B2%7Dx%20%2B%20%5Ctan%5E%7B4%7Dx%20%5D)
= Right hand side (Proved)
79. Left hand side
= 
= ![\cos x[1 - \sin^{2} x] \sin^{2} x](https://tex.z-dn.net/?f=%5Ccos%20x%5B1%20-%20%5Csin%5E%7B2%7D%20x%5D%20%5Csin%5E%7B2%7D%20x)
{Since 
}
= ![[\sin^{2}x - \sin^{4}x] \cos x](https://tex.z-dn.net/?f=%5B%5Csin%5E%7B2%7Dx%20-%20%5Csin%5E%7B4%7Dx%5D%20%5Ccos%20x)
= Right hand side
80. Left hand side
= 
= ![[\sin^{2}x + \cos^{2}x]^{2} - 2\sin^{2} x \cos^{2}x](https://tex.z-dn.net/?f=%5B%5Csin%5E%7B2%7Dx%20%2B%20%5Ccos%5E%7B2%7Dx%5D%5E%7B2%7D%20-%202%5Csin%5E%7B2%7D%20x%20%5Ccos%5E%7B2%7Dx)
{Since }
= ![1 - 2\cos^{2} x[1 - \cos^{2}x ]](https://tex.z-dn.net/?f=1%20-%202%5Ccos%5E%7B2%7D%20x%5B1%20-%20%5Ccos%5E%7B2%7Dx%20%5D)
= 
= Right hand side. (Proved)
2 and 4
Step-by-step explanation:
This image shows an example on corresponding angles.
