(8c+2a)(4+b)
(8c)(4)+(8c)(b)+(2a)(4)+(2a)(b)
32c+8bc+8a+2ab
2ab+8bc+8a+32c
A.<span>640/(120+w)
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Answer:
Q1
cos 59° = x/16
x = 16 cos 59°
x = 8.24
Q2
BC is given 23 mi
Maybe AB is needed
AB = √34² + 23² = 41 (rounded)
Q3
BC² = AB² - AC²
BC = √(37² - 12²) = 35
Q4
Let the angle is x
cos x = 19/20
x = arccos (19/20)
x = 18.2° (rounded)
Q5
See attached
Added point D and segments AD and DC to help with calculation
BC² = BD² + DC² = (AB + AD)² + DC²
Find the length of added red segments
AD = AC cos 65° = 14 cos 65° = 5.9
DC = AC sin 65° = 14 sin 65° = 12.7
Now we can find the value of BC
BC² = (19 + 5.9)² + 12.7²
BC = √781.3
BC = 28.0 yd
All calculations are rounded
Answer:
The equation has infinitely many solutions for any value of P and Q such that P=Q.
Step-by-step explanation:
Px - 37 = Qx - 37
Px - Qx - 37 = -37
x (P-Q) = 0
==> P-Q=0 ==> P=Q
