Answer:
Step-by-step explanation:
m∠ JHI + m∠GHJ = 180 { linear pair}
m∠JHI + 90 = 180
m∠JHI = 180 -90 = 90
In ΔGHJ and ΔJHI,
GJ = IJ { given}
JH = JH {common}
m∠ JHI = m∠GHJ { above proved}
ΔGHJ ≅ ΔJHI { Angle Side Side congruence}
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)
Answer:
b
Step-by-step explanation:
I did this problem before
Answer:
Step-by-step explanation:
There are 4 blue, 5 red and total of 9 marbles.
<u>If two marbles are taken and we are looking for exactly one red, then it is:</u>
- red, then blue or blue, then red
<u>Find the probability of each case:</u>
- P(r, b) = 5/9*4/8 = 5/18
- P(b, r) = 4/9*5/8 = 5/18
<u>Probability of exactly one red:</u>
