Three years after purchase, a car is estimated to be worth

1 answer:

7 0

Y - value
x - year

$24000=a\cdot3+b\\
19000=a\cdot5+b\\\\
24000=3a+b\\
-19000=-5a-b\\
---------\\
5000=-2a\\
a=-2500\\\\
b+3\cdot(-2500)=24000\\
b-7500=24000\\
b=31500\\\\
\boxed{y=-2500x+31500}

$

a)
$f(x)=-2500x+31500\\
f(x+1)=-2500(x+1)+31500\\
f(x+1)=-2500x-2500+31500\\
f(x+1)=-2500x+29000\\\\
f(x)-f(x+1)=-2500x+31500-(-2500x+29000)\\
f(x)-f(x+1)=-2500x+31500+2500x-29000\\
f(x)-f(x+1)=\boxed{2500}$

b)
$f(0)=-2500\cdot0+31500=\boxed{31500}$

c)
$f(10)=-2500\cdot10+31500\\
f(10)=-25000+31500\\
f(10)=\boxed{6500}$

You might be interested in

Determine the most specific name for the figure in the diagram.

Makovka662 [10]

<span>ABCD is a parallelogram. Looking at the quadrilateral ABCD, the first thing to do is to determine if the opposite sides are parallel to each other. So let's check that by looking at the opposite sides. Line segment BA. When you go from point B to point A, you move to the right 1 space, and down 4 spaces. So the slope is -4. Looking at line segment CD, you also move to the right 1 space and down 4 spaces, which also means a slope of -4. So those two sides are parallel. When you compare line segments BC and AD, you'll notice that for both of them, you go to the right 5 spaces and up 2 spaces, so those too are parallel. So we can now saw that the quadrilateral ABCD is a parallelogram. Since ABCD is a parallelogram, we now need to check if it's a rectangle (we know it can't be a square since the sides aren't all the same length). An easy way to test if it's a rectangle is to check of one of the angles is 90 degrees. And if we draw a line from B to D, we can create a triangle ABD. And in a right triangle, due to Pythagora's theorem we know that A^2 + B^2 = C^2 where A is the line segment AB, B is the line segment AD and C is the line segment BD. So let's calculate A^2, B^2, and C^2. A^2: Line segment AB. We can construct a right triangle with A = 1 and B = 4. So C^2 = 1^2 + 4^2 = 1 + 16 = 17. So we have an A^2 value of 17 B^2: Line segment AD. We can construct a right triangle with A = 2 and B = 5. So C^2 = 2^2 + 5^2 = 4 + 25 = 29. So we have an B^2 value of 29 C^2: Line segment BD. We can construct a right triangle with A = 2 and B = 6. So C^2 = 2^2 + 6^2 = 4 + 36 = 40. So we have a C^2 value of 40. Now let's check if the equation A^2 + B^2 = C^2 is correct: 17 + 29 = 40 46 = 40 And since 46 isn't equal to 40, that means that ABCD can not be a rectangle. So it's just a parallelogram.</span>

6 0

3 years ago

Two dice are rolled. Define the events as follows: A : sum of 7, B : doubles, C : sum of 8

love history [14]

Answer: for a sum of 7 would either be 6 and 1, 5 and 2 or 4 and 3. you can roll a double 1,2,3,4,5,6. the sum of 8 would be 2 and 6, 3 and 5 and a double 4,

Step-by-step explanation:

well the there are six sides on a die so 6 x 2 is 12 so the max sum you can get is 12. So you take 7 and you start adding up to 7, same thing for sum of 8. since there is 6 sides and a pair of dices you will roll six doubles

5 0

2 years ago

The first quartile of a data set is 18, the median is 25, the third quartile is 28,

laila [671]

The outlier would be shown as a dot so the answer is B. 45
The interquartile range is 28 - 18 = 10
A simple rule is that points more than 1.5xIRQ above or below the medians is an outlier and is shown as a point.
28 + 1.5x10 = 43. 45 is above this value.

7 0

1 year ago

Sergei says that 10% of a number will always be less than 20% of any other number.

11111nata11111 [884]

That's false.
Let's see an instance:
10% of 300=30
20% of 100=20
It doesn't work here so it's false

5 0

2 years ago

The coordinate plane below represents a city. Points A through F are schools in the city.

Usimov [2.4K]

Part A. The technique on how to find the equation that only applies to point D and E, is to create a line or curve that only includes two of these points. In this case, I created a random parabola that isolates points C and F from the rest of the points. First, we have to find the equation of the parabola through its general forms:

(x - h)² = +/-4a(y-k) or (y - k)² = +/-4a(x - h)

For parabolas drawn like that in the picture, the general form is (x - h)² = +4a(y-k), where the vertex is (h,k) and a is the distance from the vertex to the focus. From the picture, the vertex is at (0,3). Then, we use point D(-2,4) to determine a:

(-2 - 0)² = +4a(4 - 3)
4a = 4

So, the equation of the parabola is:

x² = 4(y - 3)
x² = 4y - 12

Part B. Point D was already verified above. Now for point E(2,4)
x² = 4y - 12
2² ? 4(4) - 12
4 ? 4
4 = 4

Part C. For y < 7x − 4, ignore the equality symbol first and graph the line. Assign random values of x, then you get corresponding values of y. Plot them as shown in the second picture. The line is shown in red. Next, test the equation by choosing a random point. Let's choose the purple point at (4,3).

3 ? 7(4) − 4
3 ? 24
3 < 24

Thus, it applies to the purple point, and all the other areas to that area. The shaded region are all solutions of the inequality. So, Erica is only interested in points E, C and F.

5 0

3 years ago