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Shkiper50 [21]
3 years ago
7

A new car worth 25,000 is depreciating(loses value) in value by 1,250 per year. Determine how many years will it take for the ca

r to be worth 8,500
Mathematics
1 answer:
Svet_ta [14]3 years ago
8 0
The value x years from now will be
   v(x) = 25000 - 1250x

Filling in the given information, we have
   8500 = 25000 - 1250x
   8500 - 25000 = -1250x
   -16500/-1250 = x = 13.2

It will take 13.2 years for the car to be worth 8,500.
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What is the area of this composite figure?<br>A. 135 cm2<br>B. 207 cm2<br>C. 249 cm2<br>D. 279 cm2
mariarad [96]

Answer:

249 cm^2

Step-by-step explanation:

This problem becomes easier if we subdivide the figure, find the areas of the resulting figures and then sum them up.

Draw a vertical line straight down from the edge marked "4 cm" towards the edge marked "18 cm."  The resulting rectangle on the left is 15.5 cm long and (18 - 7.5) cm wide, or 15.5 by 10.5 cm.  Its area is 162.75 cm^2.

Next, find the area of the rectangle on the right of the line we drew.  Its width is 7.5 cm and its height (15.5 - 4) cm, resulting in an area of 86.25 cm^2.

Last, add together these two subareas:  combine 86.25 cm^2 and 162.75 cm^2.  The total area of the composite figure is then 249 cm^2 (answer).

7 0
3 years ago
5.1436 rounded to the nearest hundredth
tatiyna

Answer:

5.14

Step-by-step explanation:

ones.tenths hundredths

8 0
3 years ago
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Two balls are pulled one after another, without replacement, from the box containing three black, five yellow, and seven red bal
kolezko [41]

Answer:

1/14

Step-by-step explanation:

5/15 × 3/14 = 1/14

Or 0.0714 (3sf)

6 0
3 years ago
What is 1 + 1<br><br> Please give a detailed answer and I’ll crown
Ksivusya [100]

Answer:

1+1+2

1/10x10=1

1/10x10=1

(1/10x10)+(1/10x10)+2

or...

0.5x2=1

0.5x2=1

(0.5x2)+(0.5x2)=2

hope this helps lol

6 0
2 years ago
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Given the position of the particle, what the position(s) of the particle when it’s at rest
choli [55]

The position function of a particle is given by:

X\mleft(t\mright)=\frac{2}{3}t^3-\frac{9}{2}t^2-18t

The velocity function is the derivative of the position:

\begin{gathered} V(t)=\frac{2}{3}(3t^2)-\frac{9}{2}(2t)-18 \\ \text{Simplifying:} \\ V(t)=2t^2-9t-18 \end{gathered}

The particle will be at rest when the velocity is 0, thus we solve the equation:

2t^2-9t-18=0

The coefficients of this equation are: a = 2, b = -9, c = -18

Solve by using the formula:

t=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}

Substituting:

\begin{gathered} t=\frac{9\pm\sqrt[]{81-4(2)(-18)}}{2(2)} \\ t=\frac{9\pm\sqrt[]{81+144}}{4} \\ t=\frac{9\pm\sqrt[]{225}}{4} \\ t=\frac{9\pm15}{4} \end{gathered}

We have two possible answers:

\begin{gathered} t=\frac{9+15}{4}=6 \\ t=\frac{9-15}{4}=-\frac{3}{2} \end{gathered}

We only accept the positive answer because the time cannot be negative.

Now calculate the position for t = 6:

undefined

6 0
1 year ago
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