Answer:428 7/100
Step-by-step explanation:
Answer:
1) x = 14°, y = 5°
2) x = 18.5°, y = 37°
Step-by-step explanation:
1) ∠AOC and ∠BOD are vertical angles, then ∠BOD = 25°
∠MOD = ∠MOB + ∠BOD = 90°
3x + 23° + 25° = 90°
3x = 90° - 23° - 25°
x = 42°/3
x = 14°
∠LOB = ∠LOM + ∠MOB = 90°
5y + 3x + 23° = 90°
5y = 90° - 23° - 3(14°)
y = 25°/5
y = 5°
2) ∠AOC and ∠BOD are vertical angles, then ∠BOD = 16°
∠EOB = ∠EOD + ∠DOB = 90°
2y + 16° = 90°
y = (90° - 16°)/2
y = 37°
∠DOF = ∠BOF + ∠DOB = 90°
4x + 16° = 90°
x = (90° - 16°)/4
x = 18.5°
Answer:
b(b/a)^2
Step-by-step explanation:
Given that the value of the car depreciates such that its value at the end of each year is p % less than its value at the end of the previous year and that car was worth a dollars on December 31, 2010 and was worth b dollars on December 31, 2011, then
b = a - (p% × a) = a(1-p%)
b/a = 1 - p%
p% = 1 - b/a = (a-b)/a
Let the worth of the car on December 31, 2012 be c
then
c = b - (b × p%) = b(1-p%)
Let the worth of the car on December 31, 2013 be d
then
d = c - (c × p%)
d = c(1-p%)
d = b(1-p%)(1-p%)
d = b(1-p%)^2
d = b(1- (a-b)/a)^2
d = b((a-a+b)/a)^2
d = b(b/a)^2 = b^3/a^2
The car's worth on December 31, 2013 = b(b/a)^2 = b^3/a^2
<span>3(2v + 1)= -15(5v + 16)
6v + 3 = -75v - 240 (Add 75v onto each side)
+75v +75v
81v + 3 = 240 (Subtract 3 on each side)
-3 -3
81v = -243
v = -3
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