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Solution
[tex]\begin{gathered} x^2+2x-16=0 \\ \\ \text{ using quadratic formula} \\ \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ a=1,b=2,c=-16 \\ \\ \Rightarrow x=\frac{-2\pm\sqrt{2^2-4(1)(-16)}}{2(1)} \\ \\ \Rightarrow x=\frac{-2\pm\sqrt{4+64}}{2} \\ \\ \Rightarrow x=-1-\sqrt{17} \\ \\ \Rightarrow x=-1+\sqrt{17} \\ \\ \text{ since }x>0 \\ \\ \text{ Therefore the value of }x=-1+\sqrt{17} \end{gathered}[/tex]Related Questions
Solve the quadratic equation by completing the square.x^2+6x-1=0First choose the appropriate form and fill in the blank with the correct numbets. Then, solve the equation. Round your answer to the nerest hundredth. If there is more than one solutions, separate them with commas.
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Answer:
Explanation:
Given the quadratic equation
x^2+6x-1=0
Step 1: Add 1 to both sides of the equation
x^2+6x-1 + 1 = 0 + 1
x^2 + 6x = 1
Step 2: Complete the square by adding the square of the half of coeficient of x to both sides
Coefficient of x = 6
Half of 6 = 6/2 = 3
Square of 3 = 3^2 = 9
Add 9 to both sides
x^2 + 6x + 3^2 = 1
∣/8∣=3Group of answer choicesx = 2 and x = 4x = 16 and x = 4x = -24 and x = 24x = -6 and x = -8
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Given:
[tex]|\frac{x}{8}|=3[/tex]Applying absolute value property
[tex]\frac{x}{8}=-3\text{ and }\frac{x}{8}=3[/tex]Multiply both-side by 8.
That is;
[tex]\begin{gathered} \frac{x}{8}\times8=-3\times8 \\ \\ \text{and } \\ \\ \frac{x}{8}\times8=3\times8 \end{gathered}[/tex][tex]x=-24\text{ and x=24}[/tex]Hence, x=-24 and x=24
Consider the line . y=3/2x+3Find the equation of the line that is parallel to this line and passes through the point .(-8,3)Find the equation of the line that is perpendicular to this line and passes through the point . (-8,3)
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Answer:
Equation of parallel line: y = 3x/2 + 15
Equation of perpendicular line: y = - 2x/3 - 7/3
Explanation:
The equation of a line in the slope intercept form is expressed as
y = mx + c
where
m represents slope
c represents y intercept
The equation of the given line is
y = 3x/2 + 3
By comparing with the slope intercept equation,
slope, m = 3/2
Recall, if two lines are parallel, it means that they have the same slope. Thus, the slope of the parallel line passing through the point, (- 8, 3) is 3/2. We would find the y intercept, c by substituting m = 3/2, x = - 8 and y = 3 into the slope intercept equation. We have
3 = 3/2 * - 8 + c
3 = - 12 + c
c = 3 + 12 = 15
By substituting m = 3/2 and c = 15 into the slope intercept equation, the equation of the parallel line passing through the point, (- 8, 3) is
y = 3x/2 + 15
Recall, if two lines are perpendicular, it means that the slope of one line is the negative reciprocal of the slope of the other line. Thus, the slope of the perpendicular line passing through the point, (- 8, 3) is - 2/3. We would find the y intercept, c by substituting m = - 2/3, x = - 8 and y = 3 into the slope intercept equation. We have
3 = - 2/3 * - 8 + c
3 = 16/3 + c
c = 3 - 16/3 = - 7/3
By substituting m = - 2/3 and c = - 7/3 into the slope intercept equation, the equation of the perpendicular line passing through the point, (- 8, 3) is
y = - 2x/3 - 7/3
A bag is made with 1,350 green, blue, and white beads. Twice as many green beads as blue beads are used the number of white beads is half of the total number of green and blue beads. how many green beads are used?
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Write equations for each succeeding sentences. Use G for green, B for blue, and W for white beads.
[tex]\begin{gathered} G+B+W=1350 \\ G=2B \\ W=\frac{1}{2}(G+B) \\ G=\text{?} \end{gathered}[/tex]Solve for the value of G as follows.
Rewrite the equations in terms of B. Since the value of G is already written in terms of B, write the value of W in terms of B.
[tex]\begin{gathered} W=\frac{1}{2}(G+B)_{}_{} \\ =\frac{1}{2}(2B+B) \\ =\frac{1}{2}(3B) \end{gathered}[/tex]Substitute the values of G and W, in terms of B, into the first equation and then solve for B.
[tex]\begin{gathered} G+B+W=1350 \\ 2B+B+\frac{1}{2}(3B)=1350 \\ 4B+2B+3B=2700 \\ 9B=2700 \\ B=300 \end{gathered}[/tex]Note that we obtained the third equation by multiplying both sides of the equation by 2. This eliminates the denominator, 2, from the left side of the equation.
Substitute the obtained value of B in the second given equation to solve for G.
[tex]\begin{gathered} G=2B \\ =2(300) \\ =600 \end{gathered}[/tex]Substitute the obtained value of B into the obtained value of W and then simplify.
[tex]\begin{gathered} W=\frac{1}{2}(3B) \\ =\frac{1}{2}\lbrack3(300)\rbrack \\ =\frac{1}{2}(900) \\ =450 \end{gathered}[/tex]To check if the answer is correct, add all the number of beads per color and determine if the sum is the same as the given value.
[tex]\begin{gathered} G+B+W=1350 \\ 600+300+450=1350 \\ 1350=1350 \end{gathered}[/tex]Since the equation is true, the answers are correct.
Therefore, there must be 600 green beads that were used.
The sum of the catheters in a triangle is 27 cm. The corresponding catheter in another right-angled triangle, uniform with the first one, is 2cm and 7cm. Calculate the area of the first triangle.
Answers
Given: The sum of the catheters in a triangle is 27 cm
To Determine: The area of the triangle
Solution
Please note the below
Let the first cathetus be x, then the second cathetus would be
[tex]\begin{gathered} c_1=x \\ c_2=27-x \end{gathered}[/tex]For the second right triangle
[tex]\begin{gathered} c_1=2 \\ c_2=7 \end{gathered}[/tex]Since the two right triangles are corresponding to each other, then the ratio of their cathethers are equal
Therefore
[tex]\begin{gathered} \frac{x}{27-x}=\frac{2}{7} \\ 7x=2(27-x) \\ 7x=54-2x \\ 7x+2x=54 \\ 9x=54 \\ x=\frac{54}{9} \\ x=6 \end{gathered}[/tex]So, the cathethers for the first right triangle would be
[tex]\begin{gathered} c_1=x:c_2=27-x \\ c_1=6 \\ c_2=27-6 \\ c_2=21 \end{gathered}[/tex]Note that the catheters formed the base and the height of the first triangle. The area of a triangle can be calculated using the formula below
[tex]\begin{gathered} Area(triangle)=\frac{1}{2}\times base\times height \\ Area(triangle)=\frac{1}{2}\times6cm\times21cm \\ Area(triangle)=3cm\times21cm \\ Area(triangle)=63cm^2 \end{gathered}[/tex]Hence, the area of the first triangle is 63cm²
It takes 3 1/3 spoons of chocolate syrup to make 3 1/2 į gallons of chocolate milk.How many spoons of syrup would it take to make 5 gallons of chocolate milk?
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Given:
[tex]3\frac{1}{3}spoons\text{ of chocolate syrup to make 3}\frac{1}{2}\text{ gallons of chocolate milk.}[/tex][tex]\begin{gathered} \text{Number of spoons required of syrup to make 5gallons of chocolate milk=}\frac{3\frac{1}{3}}{3\frac{1}{2}}\times5 \\ No\text{ of spoons required of syrup to make 5gallons of chocolate milk=}\frac{\frac{10}{3}}{\frac{7}{2}}\times5 \\ No\text{ of spoons required of syrup to make 5gallons of chocolate milk=}\frac{10}{3}\times\frac{2}{7}\times5 \\ No\text{ of spoons required of syrup to make 5gallons of chocolate milk=}\frac{100}{21} \\ No\text{ of spoons required of syrup to make 5gallons of chocolate milk=}4\frac{16}{21}\text{ } \end{gathered}[/tex]The diameter is 16 ftwhat's the area the circle?
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Explanation
Step 1
the area of a circle is given by:
[tex]\text{Area}_{circle}=\text{ }\pi\cdot\frac{diameter^2}{4}[/tex]let
diameter=16 ft
now, replace
[tex]\begin{gathered} \text{Area}_{circle}=\text{ }\pi\cdot\frac{diameter^2}{4} \\ \text{Area}_{circle}=\text{ }\pi\cdot\frac{(16ft)^2}{4} \\ \text{Area}_{circle}=\text{ }\pi\cdot\frac{256ft^2}{4} \\ \text{Area}_{circle}=201.06ft^2 \end{gathered}[/tex]I hope this helps you
Can anyone help me with this (there is a part two)
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Number of bottles: 6
Coupon discount on each bottle: $0.50
Final price: $5.10
If p is the regular price of each bottle, then 6p is the regular price of 6 bottles. This means that we have used 6 coupons of $0.50, so the total discount should be 6*0.50 dollars. We subtract this amount from the regular price (6p), leading to $5.10. The equation that represents this situation is:
[tex]\begin{gathered} 6p-6\cdot0.50=5.10 \\ \Rightarrow6(p-0.50)=5.10 \end{gathered}[/tex]Given the area of triangle AEC=63cm^2, find the area of triangle ABC.
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We are given that the area of triangle AEC = 63 centimeters squared.
Since segment CD equals segment DB that means that triangle CDA and triangle BDA have the same area, and also triangle CDE and triangle BDE have the same area. This means mathematically the following:
[tex]A_{\text{ADC}}-A_{\text{AEC}}=A_{\text{ADB}}-A_{\text{AEB}},\text{ (1)}[/tex]Also
[tex]A_{\text{ADC}}=A_{\text{ADB}},\text{ (2)}[/tex]Replacing equation (1) in equation (2)
[tex]A_{\text{ADC}}-A_{\text{AEC}}=A_{\text{ADC}}-A_{\text{AEB}}[/tex]Simplifying
[tex]A_{\text{AEC}}=A_{\text{AEB}}[/tex]Therefore:
[tex]A_{\text{AEB}}=63\operatorname{cm}^2[/tex]Since segments DE and EA is the same, then:
[tex]A_{\text{CDE}}=A_{\text{AEC}}[/tex]Therefore:
[tex]A_{\text{CDE}}=63\operatorname{cm}^2[/tex]Since
[tex]A_{\text{CDE}}=A_{\text{BDE}}[/tex]We have:
[tex]A_{\text{BDE}}=63\operatorname{cm}^2[/tex]therefore, the area of the triangle is:
[tex]A_{\text{ABC}}=A_{\text{AEC}}+A_{\text{AEB}}+A_{\text{CDE}}+A_{\text{BDE}}[/tex]Replacing the known values:
[tex]\begin{gathered} A_{\text{ABC}}=68+68+68+68=4(68) \\ A_{\text{ABC}}=272\operatorname{cm}^2 \end{gathered}[/tex]Which expression has the same value as sin(20∘)?A)cos(10∘)B)cos(20∘)C)cos(40∘)D)cos(70∘)
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sin 20 = y / r
cos 70 = y/ r
sin 20 = cos 70 = 0.342
Answer:
cos 70
LEARNING OBJECTIVE Determine a vertical Horizontal or oblique asymptole of a rational functionWhich of the following rational functions will have a graph with a horizontal asymptote of y=09nh7x) =2x + 4b.)4x) - 2x + 23x - 1c.)3x - 2x2x+X-1d.)2x - 83x+x+1
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We need to find a vertical asymptote. This means, when we are approaching a value X, then Y becomes infinite or -infinite
A rational function R(x) = p(x) / q(x) will have a vertical asymptote at x=r when r is substituted in for x it makes the denominator zero but not the numerator
option a) oblique asymptote
option b) we have both horizontal (at y=0) and vertical (at x=-1) asymptotes
Option c)
option d) horizontal asymptote
I need help with my math
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The Slope of a Line
Suppose we know the line passes through points A(x1,y1) and B(x2,y2). The slope can be calculated with the equation:
[tex]\displaystyle m=\frac{y_2-y_1}{x_2-x_1}[/tex]The graph provided suggests the use of the points (3,-3) and (5,-3). The slope is:
[tex]\displaystyle m=\frac{-3+3}{5-3}=\frac{0}{2}=0[/tex]The slope of the line is 0. It corresponds to a horizontal line
2. You pay $18.00 for 30 text messages. At the same rate, how much would 12text messages cost?17
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$18 for 30 messages
Ratio = price / messages = 18/30
For 12 messages:
Price / messages = x /12
Equal both ratios:
18/30 = x /12
Solve for x:
0.6 (12) = x
$7.2 = x
$7.2 for 12 messages
1.X: -2, -1, 0, 1, 2Y: -7, -2, 1, -2, -7Domain:Range:Function: Yes Or no?
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We have
X: -2, -1, 0, 1, 2
Y: -7, -2, 1, -2, -7
the domain is the set of all the possible values for x, in this case, we have
{-2, -1, 0, 1, 2}
the range is the set of all possible values of y in this case we have
{-7, -2, 1}
With this information, we can say it is a function,
Write an equation parallel to y = 3x + 6 that passes through the point (4,7).Remember to type the" - "if a number is negative, such as-2.y =X +
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The equation that is parallel to y = 3x + 6 has the same slope as y, namely 3; therefore, we already know that the equation we are seeking has the form
[tex]y=3x+b[/tex]Now we just need to solve for the y-intercept b, and to do that we use the point (4, 7 ). Putting x = 4 and y = 7 into the above equation gives
[tex]7=3(4)+b[/tex][tex]7=12+b[/tex][tex]\therefore b=-5[/tex]Hence, the equation that is parallel to y = 3x + 6 that passes through the point (4,7) is
[tex]y=3x-5[/tex]
Using the following images, name the intersection of line QS and line LC.
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The intersection between two non-parallel lines is a point, as we can see in the following diagram:
As we can see from the image, the intersection point between QS and LC is W.
Answer: Point W
What value of x will make the following equation true? Log4(4^5x+1)=16•7/5•0•1/5•3(Picture for clarification)
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Given:
There are given that expression:
[tex]log_4\left(4^{5x+1}\right?=16[/tex]Explanation:
From the given log function:
[tex]log_4(4^{5x+1})=16[/tex]According to the log rule:
[tex]5x+1=16[/tex]Then,
[tex]\begin{gathered} 5x+1=16 \\ 5x=16-1 \\ x=\frac{15}{5} \\ x=3 \end{gathered}[/tex]Final answer:
Hence, the correct option is D.
Time(wki469Height ofplant (in)9.013.520.25Find the rate of change for weeks 40le and 69.Explain the meaning of the rate of change for each case.
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Explanation
Step 1
when you have 2 coordinates ( A and B), the slope of the line that passes thought those point is given by
[tex]\text{slope}=\frac{y_2-y_1}{x_2-x_1}[/tex]where
[tex]\begin{gathered} A=(x_1,y_1) \\ B=(x_2,y_2) \end{gathered}[/tex]A and B are 2 known points of the line
Step 2
so, the slope represents the rate of change
i)the rate of change for 4-6 weeks
Let
A=(4 ,9)
B=(6,13.5)
replace
[tex]\begin{gathered} \text{slope}=\frac{y_2-y_1}{x_2-x_1} \\ slope_1=\frac{13.5-9}{6-4}=\frac{4.5}{2}=2.25 \\ slope_1=2.25 \end{gathered}[/tex]Step 3
ii)the rate of change for 6-9weeks
Let
A(6,13.5)
B(9,20.25)
replace,
[tex]\begin{gathered} \text{slope}=\frac{y_2-y_1}{x_2-x_1} \\ \text{slope}=\frac{20.25-13.5}{9-6}=\frac{6.75}{3}=2.25 \end{gathered}[/tex]the slope represents the rate of change, it means for every case the plant is growing at a constant rate (2.25 inches per week)
I hope this helps you
Line A is perpendicular to Line B.If the slope of Line A is -5,what is the slope of Line B?391410
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Given:
Line A is perpendicular to Line B.
Required:
what is the slope of Line B?
Explanation:
Based on the given conditions, formulate.
[tex]m=\frac{-5}{3}[/tex]Find the slope of line that is perpendicular to
[tex]\frac{-5}{3}[/tex][tex]m=\frac{3}{5}[/tex]Required answer:
[tex]m=\frac{3}{5}[/tex]Managers of a sports arena’s parking garage keep track of the duration of time customers park their cars there. Shown in the stem and - leaf display below is a sample of 15 such parking duration (in minutes). Use the display to answer the questions that follow.
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Step 1
A Stem and Leaf Plot is a special table where each data value is split into a "stem" (the first digit or digits) and a "leaf" (usually the last digit).
[tex]198\text{ Minutes}[/tex]Step 2
[tex]\begin{gathered} In\text{ the 180s, we have;} \\ 182,183,186,189\text{ minutes} \\ The\text{ shortest parking duration in the 180's is 182} \\ Answer=182\text{ Minutes} \end{gathered}[/tex]Step 3
[tex]\begin{gathered} In\text{ the 160's, we have; 160,164,164} \\ Answer=3\text{ } \end{gathered}[/tex]Wich function is used to find y,the remaining balance after x number of payments have been made?
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The slope of a line that passes through points (x1, y1) and (x2, y2) is computed as follows:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]The line of the picture passes through (0, 20000) and (40, 0) then its slope is:
[tex]m=\frac{0-20000}{40-0}=-500[/tex]The y-intercept of the line is (0, 20000)
Slope-intercept form of a line:
y = mx + b
where m is the slope and b is the y-coordinate of the y-intercept. Replacing with m = -500 and b = 20000, we get:
y = -500x + 20000
When a figure is translated its orientation (blank) and the measurements of its angles (blank).The options for both blanks are the same and the options are, remain the same or change
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First of all, remember that translation is a transformation which doesn't imply a change of size or shape, that is, the image will be congruent to its image.
Having said that, the complete paragraph would be
When a figure is translated its orientation remains the same and the measurements of its angles remain the same.
The orientation doesn't change because it's defined as the position of points of the figure, these points change its position where we rotate the figure, which is not the case here.
if like bc is parallel to line AD what is the measure of BAD
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D.
if anyone could help me on #17 i would appreciate it!
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Answer:
[tex]f(x)=-\lvert x-7\rvert+2[/tex]Step-by-step explanation:
The function that was transformed is:
[tex]f(x)=\lvert x\rvert[/tex]If it reflects in the x-axis, shift 7 units to the right, and shift upward 2 units, we need to know the transformation rules for these displacements:
[tex]\begin{gathered} \text{ -f(x) reflects the function in the x-axis (upside-down)} \\ f(x-b)\text{ shifts the function b units to the right.} \\ \text{ f(x)+b shifts the function b units upward.} \end{gathered}[/tex]Now, with this in mind, the equation of the function transformed would be:
[tex]f(x)=-\lvert x-7\rvert+2[/tex]What is anequation of the line that passes through the points (-7, -7) and(-7,4)?
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The line passes through (-7,-7) and (-7,4); thus the x-component remains fixed but the y-component is free
the y- component can take any value! thus the equation is
x=-7
The value of an antique car is modeled by the function
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when we are modeling increments using functions the standard form should be
[tex]V(t)=A\cdot(1+r)^t[/tex]In which A represents the initial value and r represents the rate it is increasing per year.
In this case to find what is the increment per year we equal what is inside the parentheses
[tex]\begin{gathered} 1+r=1.004 \\ r=1-1.004 \\ r=0.004 \end{gathered}[/tex]now this decimal can be represented as a percentage if we multiply by 100
[tex]\begin{gathered} \text{\%r}=0.004\cdot100 \\ \text{\%r=0.4\%} \end{gathered}[/tex]It is increasing by 0.4% per year.
Find the quotient. Express the final result using positive integer exponents only (72x^-1 y^4)^-1 / (8x^8y^3)
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ANSWER:
[tex]\frac{1}{576x^7y^7}[/tex]STEP-BY-STEP EXPLANATION:
We have the following function:
[tex]\frac{\left(72x^{-1}y^4\right)^{-1}}{8x^8y^3}[/tex]We operate to simplify and we are left with the following:
[tex]\frac{72^{-1}x^{-1\cdot-1}y^{4\cdot-1}}{8x^8y^3}=\frac{\frac{1}{72}xy^{-4}}{8x^8y^3}=\frac{1}{72\cdot8\cdot x^8\cdot x^{-1}\cdot y^3\cdot y^4}=\frac{1}{576x^7y^7}[/tex]Chris rented a truck for one dah There was a base fee of 18.95$ and there was an additional charge of 83 cents for each mile driven. Chris had to pay 209.02 when he returned the truck. For how many miles did he drive the truck?
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Given:
Base fee $18.95
83 cents for each mile driven.
[tex]\text{Amount excluding the base fee=209.02-18.95}[/tex][tex]\text{Amount excluding the base fee= \$}190.07[/tex][tex]\text{Number of miles driven =}\frac{\text{19007}}{83}[/tex][tex]undefined[/tex]When Ryan runs the 400 meter dash, his finishing times are normally distributedwith a mean of 65 seconds and a standard deviation of 2 seconds. If Ryan were to run.36 practice trials of the 400 meter dash, how many of those trials would be between63 and 65 seconds, to the nearest whole number?
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Solution
Step 1
Dalvin's finishing time is normally distributed with a mean of 65 seconds and a standard deviation of 1 second.
[tex]\begin{gathered} \text{Mean }\mu\text{ = 65} \\ Standard\text{ deviation }\sigma\text{ = 1} \end{gathered}[/tex]Step 2
Under the empirical rule, 68% of the results will be within 1 standard deviation.
Step 3
Since the standard deviation is 1 second, 68% of Dalvin's finishing time will be between 63 and 65 seconds.
Final answer
68%
The equation V=15200(0.93) t V=15200 (0.93)t represents the value (in dollars) of a car t years after its purchase
Answers
We will have the following:
[We can see a constant compound of the decrease in price]
The value of this car is decreasing at a rate of 7 percent.
The purchase price of the car was 15 200 dollars.