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san4es73 [151]
2 years ago
13

The selling price of a car is $13,500. Each year, it loses 11% of its value. Find the exponential function that gives the value

of the car t years after its purchase. Write your answer in the form f(t)=a(b)t.
Mathematics
1 answer:
Katen [24]2 years ago
4 0

Answer:

f(t) = 13,500(0.89)^t

Step-by-step explanation:

The formula for the exponential function is given as f(t) = a(b)^t

Where,

base value = a = $13,500

b = 1 - % decrease = 1 - 11% = 1 - 0.11

b = 0.89

Substitute the values of a and b into the exponential function, f(t) = a(b)^t, respectively.

Thus:

f(t) = 13,500(0.89)^t

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1.86 Miles per hour

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Ken went to the store for his mother. He spent $2.37 on groceries. His mother said he could spend 35¢ on candy. What was his cha
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Jose is raising money for a school trip by selling lollipops and packs of cookies. The price of each lollipop is $1 and the pric
Ratling [72]

Answer:

For the first question, he would raise $38.

Step-by-step explanation:

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4 0
3 years ago
Please help with my math
shusha [124]

Answer:

Step-by-step explanation:

These triangles are similar triangles, so there is a number that you can multiply the sides of TUV to find the side lengths of QRS. looking at the triangle, the similar sides are RS being similar to UV and RQ being similar to UT.

If RS~UV, then there is a ratio between them. 54/36=1.5. The ratio is 1.5.

RQ~UT, and by a factor of 1.5, so divide RQ by the scale factor. 24/1.5=16. UT=16=x+5.

x+5=16, subtract 5 from both sides.

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5 0
2 years ago
3. The curve C with equation y=f(x) is such that, dy/dx = 3x^2 + 4x +k
Andreas93 [3]

a. Given that y = f(x) and f(0) = -2, by the fundamental theorem of calculus we have

\displaystyle \frac{dy}{dx} = 3x^2 + 4x + k \implies y = f(0) + \int_0^x (3t^2+4t+k) \, dt

Evaluate the integral to solve for y :

\displaystyle y = -2 + \int_0^x (3t^2+4t+k) \, dt

\displaystyle y = -2 + (t^3+2t^2+kt)\bigg|_0^x

\displaystyle y = x^3+2x^2+kx - 2

Use the other known value, f(2) = 18, to solve for k :

18 = 2^3 + 2\times2^2+2k - 2 \implies \boxed{k = 2}

Then the curve C has equation

\boxed{y = x^3 + 2x^2 + 2x - 2}

b. Any tangent to the curve C at a point (a, f(a)) has slope equal to the derivative of y at that point:

\dfrac{dy}{dx}\bigg|_{x=a} = 3a^2 + 4a + 2

The slope of the given tangent line y=x-2 is 1. Solve for a :

3a^2 + 4a + 2 = 1 \implies 3a^2 + 4a + 1 = (3a+1)(a+1)=0 \implies a = -\dfrac13 \text{ or }a = -1

so we know there exists a tangent to C with slope 1. When x = -1/3, we have y = f(-1/3) = -67/27; when x = -1, we have y = f(-1) = -3. This means the tangent line must meet C at either (-1/3, -67/27) or (-1, -3).

Decide which of these points is correct:

x - 2 = x^3 + 2x^2 + 2x - 2 \implies x^3 + 2x^2 + x = x(x+1)^2=0 \implies x=0 \text{ or } x = -1

So, the point of contact between the tangent line and C is (-1, -3).

7 0
2 years ago
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