We solve the inequality by subtracting 56.50 from both sides of the equation,
10.45b + 56.50 - 56.50 < 292.67 - 56.50
10.45b < 236.17
Then, divide both sides of the inequality by 10.45
b < 22.6
The solution suggests that the number of boxes than can be loaded on a truck without exceeding the weight limit of the truck should always be lesser than 22.6. Since we are talking about number of boxes, the maximum number of boxes that can be loaded should only be 22.
Answer:
Step-by-step explanation:
12.9 + 10.2x = 35.1
10.2x = 35.1 - 12.9
10.2x = 22.2
x = 22.2/10.2
x= 222/102 = 111/51
4500 is both divisible by 5 and 9
4500/9=500
4500/5=900
Group together the like terms (plus 2x with 2x) (subtract 10 by 6)
4x+4=36
Move plus four to the other side as a negative
4x=32
Divide both sides by 4
32 divided by 4 is 8
Therefore, x is equal to 8
Write the left side of the given expression as N/D, where
N = sinA - sin3A + sin5A - sin7A
D = cosA - cos3A - cos5A + cos7A
Therefore we want to show that N/D = cot2A.
We shall use these identities:
sin x - sin y = 2cos((x+y)/2)*sin((x-y)/2)
cos x - cos y = -2sin((x+y)/2)*sin((x-y)2)
N = -(sin7A - sinA) + sin5A - sin3A
= -2cos4A*sin3A + 2cos4A*sinA
= 2cos4A(sinA - sin3A)
= 2cos4A*2cos(2A)sin(-A)
= -4cos4A*cos2A*sinA
D = cos7A + cosA - (cos5A + cos3A)
= 2cos4A*cos3A - 2cos4A*cosA
= 2cos4A(cos3A - cosA)
= 2cos4A*(-2)sin2A*sinA
= -4cos4A*sin2A*sinA
Therefore
N/D = [-4cos4A*cos2A*sinA]/[-4cos4A*sin2A*sinA]
= cos2A/sin2A
= cot2A
This verifies the identity.
