A1, B2, L3
A1, L2, B3
L1, B2, A3
L1, A2, B3
B1, A2, L3
B1, L2, A3
Answer: see proof below
<u>Step-by-step explanation:</u>
Use the following Product to Sum Identities:
2 sin A sin B = cos (A - B) - cos (A + B)
2 sin A cos B = sin (A + B) + sin (A - B)
Use the Unit Circle to evaluate: cos 120 = -1/2 & sin 60 = √3/2
<u>Proof LHS → RHS</u>
LHS: sin 20 · sin 40 · sin 80
Regroup: (1/2) sin 20 · 2 sin 40 · sin 80
Product to Sum Identity: (1/2) sin 20 [cos(80-40) - cos (80+40)]
Simplify: (1/2) sin 20 [cos 40 - cos 120]
Unit Circle: (1/2) sin 20 [cos 40 + (1/2)]
Distribute: (1/2) sin 20 cos 40 + (1/4) sin 20
Product to Sum Identity: (1/4)[sin(20 + 40) + sin (20 - 40)] + (1/4) sin 20
Simplify: (1/4)[sin 60 + sin (-20)] + (1/4) sin 20
= (1/4)[sin 60 - sin 20] + (1/4) sin 20
Unit Circle: (1/4)[(√3/2) - sin 20] + (1/4) sin 20
Distribute: (√3/8) - (1/4) sin 20 + (1/4) sin 20
Simplify: √3/8
LHS = RHS: √3/8 = √3/8 ![\checkmark](https://tex.z-dn.net/?f=%5Ccheckmark)
Answer:
81
Step-by-step explanation:
u multiply 9 x 9 not 9x2
Answer: C) Sometimes positive; sometimes negative
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Explanation:
Pick a value between x = -1 and x = 0. Let's say we go for x = -0.5
Plug this into f(x)
f(x) = x(x+3)(x+1)(x-4)
f(-0.5) = -0.5(-0.5+3)(-0.5+1)(-0.5-4)
f(-0.5) = -0.5(2.5)(0.5)(-4.5)
f(-0.5) = 2.8125
We get a positive value.
This shows that f(x) is positive on the region of -1 < x < 0
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Now pick a value between x = 0 and x = 4. I'll use x = 1
f(x) = x(x+3)(x+1)(x-4)
f(1) = 1(1+3)(1+1)(1-4)
f(1) = 1(4)(2)(-3)
f(1) = -24
Therefore, f(x) is negative on the interval 0 < x < 4
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In short, f(x) is both positive and negative on the interval -1 < x < 4
It's positive when -1 < x < 0
And it's negative when 0 < x < 4
Answer:10
Step-by-step explanation: