We can start by finding the gradient of LM

Two perpendicular lines will meet the requirement

×

=-1
Two parallel lines have equal gradients
NM is perpendicular to LM, hence the gradient of NM is -1
KN is a line that is parallel to NM, hence the gradient is 1
KL is perpendicular to LM, hence the gradient of KL is -1
Answer: $24000
Step-by-step explanation:
Depreciation for 2018 = ($74,000 - $18000) / 7
= $56000/7
= $8000
Depreciation for 2019 = ($74,000 - $18000) / 7
= $56000/7
= $8000
Depreciation up to 2019 = $8000 + $8000 = $16000
Book value at end of 2019 = $74000 - $16000 = $58000
Revised residual value = $10,000
Number of remaining years = 4-2 = 2 years
Depreciation expensed for 2020 will be:
= ($58000 - $10,000) / 2
= $48000/2
= $24000
Think of factor pairs. She could buy 1 pair for $54, 54 pairs for $1, 2 pairs for $27, 27 pairs for $2, 3 pairs for $18, 18 for $3, 6 pairs for $9, or 9 pairs for $6
Answer:
Option d. $22154 is the right answer.
Step-by-step explanation:
To solve this question we will use the formula 
In this formula A = amount after time t
P = principal amount
r = rate of interest
n = number of times interest gets compounded in a year
t = time
Now Lou has principal amount on the starting of first year = 10000+5000 = $15000
So for one year 

= $15900
After one year Lou added $5000 in this amount and we have to calculate the final amount he got
Now principal amount becomes $15900 + $ 5000 = $20900
Then putting the values again in the formula



So the final amount will be $22154.