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Bad White [126]
3 years ago
11

An equation for the depreciation of a car is given by y = A(1 - 1)', where y = current value of the car, A = original cost, r =

rate
of depreciation, and t = time, in years. The current value of a car is $12,282.50. The car originally cost $20,000 and
depreciates at a rate of 15% per year. How old is the car?

2 years
3 years
3 years
4 years
Mathematics
1 answer:
bogdanovich [222]3 years ago
3 0
I’m pretty sure the answer is 3years
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(5x^4 - 3x^3 + 6x) - ( 3x^3 + 11x^2 - 8x)
sergeinik [125]
(5x⁴ - 3x³ + 6x) - (3x³ + 11x² - 8x)<span>

</span>Expand the second bracket by multiplying throughout by -1
5x⁴ - 3x³ + 6x - <span>3x³ - 11x² + 8x
</span>
Group like terms and simplify 
5x⁴ - 3x³ - 3x³ - 11x² + 6x <span>+ 8x

</span>5x⁴ <span>- 6x</span>³ - <span>11x² + 14x</span>

3 0
3 years ago
(−4)−(−2)–{(−5)–[(−7)+(−3)–(−8)]}
garik1379 [7]

Answer:

1

Step-by-step explanation:

(−4)−(−2)–{(−5)–[(−7)+(−3)–(−8)]}

-4 + 2 - {-5 - [-7 - 3 + 8]}

-2 - [-5 + 7 + 3 - 8]

-2 - (-3)

-2 + 3

1

7 0
3 years ago
What is the solution of this proportion 4/9 = 6/d
Alchen [17]

Answer:

d = 2/27

Remember that when a variable is times with a number you <u>divide</u> the number on both sides.

When a variable is divided by a number, you <u>multiply</u> the number on both sides.

Hope this helps you and thank you !!

4 0
3 years ago
Please help <br><br>1 through 3​
Wewaii [24]

Answer:

  1. object is moving to the right with constant speed
  2. object is moving to the left with constant speed
  3. object was stationary for a while, then started moving to the right with constant speed

Step-by-step explanation:

These graphs are of position, so the slope of the graph is the change of position with time, which is velocity. When the slope is positive, the velocity is positive, meaning its direction is to the right. When the slope is negative, the velocity is negative, meaning its direction is to the left.

When the slope is zero, the object is stationary (not moving). The position remains as it was.

1. The position vs. time curve is a straight line with positive slope. The object is moving to the right with constant velocity.

__

2. The position vs. time curve is a straight line with negative slope. The object is moving to the left with constant velocity.

__

3. The position vs. time curve is flat for a while, then increasing with constant slope. The object stayed where it was for a while, then began moving to the right (to larger values of x) with constant velocity.

7 0
3 years ago
If kenny earns 142% of Benny's Salary, how mucho more does Kenny earn than Benny
Sergeeva-Olga [200]

Let  Benny's salary = $ K

But kenny earns 142% of Benny's Salary

So, Kenny's salary = 142%of K

Kenny's salary =K  \times \frac{142}{100}

Kenny's salary = 1.42K

Kenny's income exceeds Benny's by =1.42K - K =$0.42 K

% by which Kenny's salary exceeds by Benny's =\frac{0.42 \text K}{\text K} \times 100

% by which Kenny's salary exceeds by Benny's = 42%


8 0
3 years ago
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