The value of the truck initially, Ao is
83000
1-0.16=0.84
1-0.26=0.74
After one year the value
Y=83,000×(0.84)=69,720
Y=83,000×(0.74)=61,420
When you compare the results you will see that the graph would fall at a faster rate to the right because the depreciation rate of 26% is higher than the depreciation rate of 16%
Hope it helps
Answer:
$2000 was invested at 5% and $5000 was invested at 8%.
Step-by-step explanation:
Assuming the interest is simple interest.
<u>Simple Interest Formula</u>
I = Prt
where:
- I = interest earned.
- P = principal invested.
- r = interest rate (in decimal form).
- t = time (in years).
Given:
- Total P = $7000
- P₁ = principal invested at 5%
- P₂ = principal invested at 8%
- Total interest = $500
- r₁ = 5% = 0.05
- r₂ = 8% = 0.08
- t = 1 year
Create two equations from the given information:


Rewrite Equation 1 to make P₁ the subject:

Substitute this into Equation 2 and solve for P₂:





Substitute the found value of P₂ into Equation 1 and solve for P₁:



$2000 was invested at 5% and $5000 was invested at 8%.
Learn more about simple interest here:
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Answer: Im not sure
Step-by-step explanation:
you didnt list/provide the graphs for me to answer. :I
Answer:
1.23d = 52
Explanation:
If 52 meters is the water level after a 23% increase, then we can say that the initial depth d added to the 23% of d is equal to 52 meters. So:
d + 23%d = 52 meters
Since 23% is equivalent to 0.23, we get:
d + 0.23d = 52
Finally, adding the like terms, we get:
(1 + 0.23)d = 52
1.23d = 52
So, the equation is:
1.23d = 52
Answer:
D
Step-by-step explanation:
you take away 12 from the total and you'll find how many Zach read (20) then 12+20=32