How to draw perpendicular line to a line through given point

F(x) = 15,000(.84) shows the value of a car each year. Describe the following:

makvit [3.9K]

Answers:

a) 15000 represents the starting amount
b) The decay rate is 16%, which means the car loses 16% of its value each year.
c) x is the number of years
d) f(x) is the value of the car after x years have gone by

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Explanation:

We have the function f(x) = 15000(0.84)^x. If we plug in x = 0, then we get,

f(x) = 15000(0.84)^x

f(0) = 15000(0.84)^0

f(0) = 15000(1)

f(0) = 15000

In the third step, I used the idea that any nonzero value to the power of 0 is always 1. The rule is x^0 = 1 for any nonzero x.

So that's how we get the initial value of the car. The car started off at $15,000.

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The growth or decay rate depends entirely on the base of the exponential, which is 0.84; compare it to 1+r and we see that 1+r = 0.84 solves to r = -0.16 which converts to -16%. The negative indicates the value is going down each year. So we have 16% decay or the value is going down 16% per year.

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The value of x is the number of years. In the first section, x = 0 represented year 0 or the starting year. If x = 1, then one full year has passed by. For x = 2, we have two full years pass by, and so on.

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The value of f(x) is the value of the car after x years have gone by. We found that f(x) = 15000 when x = 0. In other words, at the start the car is worth $15,000. Plugging in other x values leads to other f(x) values. For example, if x = 2, then you should find that f(x) = 10584. This means the car is worth $10,584 after two years.

7 0

3 years ago

Solve and check <br> please please help i will mark u as a brill it is easy please

Lesechka [4]

Answer:

B=25

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Expand (2x+2)^6<br> How would you find the answer using the binomial theorem?

Yanka [14]

Answer:

Step-by-step explanation:

$\displaystyle\\\sum\limits _{k=0}^n\frac{n!}{k!*(n-k)!}a^{n-k}b^k .\\\\k=0\\\frac{n!}{0!*(n-0)!}a^{n-0}b^0=C_n^0a^n*1=C_n^0a^n.\\\\ k=1\\\frac{n!}{1!*(n-1)!} a^{n-1}b^1=C_n^1a^{n-1}b^1.\\\\k=2\\\frac{n!}{2!*(n-2)!} a^{n-2}b^2=C_n^2a^{n-2}b^2.\\\\k=n\\\frac{n!}{n!*(n-n)!} a^{n-n}b^n=C_n^na^0b^n=C_n^nb^n.\\\\C_n^0a^n+C_n^1a^{n-1}b^1+C_n^2a^{n-2}b^2+...+C_n^nb^n=(a+b)^n.$

$\displaystyle\\(2x+2)^6=\frac{6!}{(6-0)!*0!} (2x)^62^0+\frac{6!}{(6-1)!*1!} (2x)^{6-1}2^1+\frac{6!}{(6-2)!*2!}(2x)^{6-2}2^2+\\\\ +\frac{6!}{(6-3)!*3!} (2a)^{6-3}2^3+\frac{6!}{(6-4)*4!} (2x)^{6-4}b^4+\frac{6!}{(6-5)!*5!}(2x)^{6-5} b^5+\frac{6!}{(6-6)!*6!}(2x)^{6-6}b^6. \\\\$

$(2x+2)^6=\frac{6!}{6!*1} 2^6*x^6*1+\frac{5!*6}{5!*1}2^5*x^5*2+\\\\+\frac{4!*5*6}{4!*1*2}2^4*x^4*2^2+ \frac{3!*4*5*6}{3!*1*2*3} 2^3*x^3*2^3+\frac{4!*5*6}{2!*4!}2^2*x^2*2^4+\\\\+\frac{5!*6}{1!*5!} 2^1*x^1*2^5+\frac{6!}{0!*6!} x^02^6\\\\(2x+2)^6=64x^6+384x^5+960x^4+1280x^3+960x^2+384x+64.$

8 0

1 year ago

Question

Mashutka [201]

The equivalent value of given the expression is -17.25. Therefore, option C is the correct answer.

<h3>What is an equivalent expression?</h3>

Equivalent expressions are expressions that work the same even though they look different. If two algebraic expressions are equivalent, then the two expressions have the same value when we plug in the same value for the variable.

The given expression is -14-8×0.5+0.75.

Now, -14-8×0.5+0.75

= -14-(8×0.5)+0.75

= -14-4+0.75

= -18+0.75

= -17.25

The equivalent value of given the expression is -17.25. Therefore, option C is the correct answer.

To learn more about an equivalent expression visit:

brainly.com/question/28170201.

#SPJ2

4 0

1 year ago