Answer:
To the nearest hundred dollars, the car will be worth $17,900 by 2005
Step-by-step explanation:
Firstly, we need to write the depreciation equation
We have this as:
V = I(1 - r)^t
V is the present value which is what we want to calculate
I is the initial value, the amount the cad was bought which is $22,000
r is the rate of change which is 5% = 5/100 = 0.05
t is the time difference which is 2005-2001 = 4
Substituting all these into the depreciation equation, we have it that
V = 22,000(1 -0.05)^4
V = $17,919.1375
To the nearest hundred dollars, that would be;
$17,900
Answer:
13
Step-by-step explanation:
Explanation:
Use the absolute value definition to rewrite the absolute value equation as two separate equations.
k-10=3
k-10=-3
Solve the equation for k
k=13
k-10=-3
k=13
k=7
The equation has 2 solutions, meaning k1=7,k2=13. Which gives you the answer of: 13 (K1=7,k2=13)
Let x = # of tickets sold in advance
Let y = # of tickets sold the day of
Cost of the tickets & total sales: 6x & 10y = 6828
You also know y = x + 206
Take the equation mentioned above y = x + 206 and sub it in anywhere the variable y is in the other equations so you'll have this:
6x + 10(x+206) = 6828
Now solve for x to get x = 298
To finish the problem, you must now find the number of y tickets sold.
Sub your x value that you found back into the equation y = x + 206 and you'll get y = 504.
So, 298 tickets were sold in advance and 504 tickets were sold the day of
Using the scale ratio, the length of the dress is: 6 ft.
<h3>How to Apply Scale to Measurement?</h3>
Given a scale for the dress pattern as, 2 inches : 1 foot, it means 2 inches of the model equals 1 foot of the actual dress.
Therefore, using the scale ratio, let x represent the length of the dress. We would have:
2/1 = 12/x
x = (12 × 1)/2
x = 6 ft.
The length of the dress is: 6 ft.
Learn more about scale on:
brainly.com/question/2429309
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Option A is correct as if we solve the equation We will find that the above equation can be compared to the standard quadratic equation i.e ax^2+bx+c=0
