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irina1246 [14]
3 years ago
5

Mandy bought a desktop computer system to start her business from home for $4,995. It is expected to depreciate at a rate of 10%

per year. After how many years will the value of her home computer system depreciates to approximately $2150
Mathematics
1 answer:
ziro4ka [17]3 years ago
3 0

Answer:

8.43 years

Step-by-step explanation:

Given data

Cost price P= $4995

Rate r= 10%

FInal amount A= $2150

The expression for the time/duration is given as

t= ln(A/p)/r

t= ln(2150/4995)/0.1

t= ln(0.430)/0.1

t= 0.843/0.1

t= 8.43

Hence the time it will take is 8.43 years

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Part 3 - Discussion/Explanation Question
SpyIntel [72]

Step-by-step explanation:

Vertical asymptote can be Identites if there is a factor only in the denominator. This means that the function will be infinitely discounted at that point.

For example,

\frac{1}{x - 5}

Set the expression in the denominator equal to 0, because you can't divide by 0.

x - 5 = 0

x = 5

So the vertical asymptote is x=5.

Disclaimer if you see something like this

\frac{(x - 5)(x + 3)}{(x - 5)}

x=5 won't be a vertical asymptote, it will be a hole because it in the numerator and denominator.

Horizontal:

If we have a function like this

\frac{1}{x}

We can determine what happens to the y values as x gets bigger, as x gets bigger, we will get smaller answers for y values. The y values will get closer to 0 but never reach it.

Remember a constant can be represent by

a \times  {x}^{0}

For example,

1 = 1 \times  {x}^{0}

2 =  2 \times {x}^{0}

And so on,

and

x =  {x}^{1}

So our equation is basically

\frac{1 \times  {x}^{0} }{ {x}^{1} }

Look at the degrees, since the numerator has a smaller degree than the denominator, the denominator will grow larger than the numerator as x gets larger, so since the larger number is the denominator, our y values will approach 0.

So anytime, the degree of the numerator < denominator, the horizontal asymptote is x=0.

Consider the function

\frac{3 {x}^{2} }{ {x}^{2}  + 1}

As x get larger, the only thing that will matter will be the leading coefficient of the leading degree term. So as x approach infinity and negative infinity, the horizontal asymptote will the numerator of the leading coefficient/ the leading coefficient of the denominator

So in this case,

x =  \frac{3}{1}

Finally, if the numerator has a greater degree than denominator, the value of horizontal asymptote will be larger and larger such there would be no horizontal asymptote instead of a oblique asymptote.

8 0
2 years ago
If (ax+2)(x+b) = 3x^2 + 5x + c for all values of x, what is the value of c?
mr Goodwill [35]

Answer: is it 0.258

Step-by-step explanation:

4 0
3 years ago
The perpendicular line of x-6y=2, (2, 4) in slope intercept form
pentagon [3]
The perpendicular line to x-6y=2, and passing through (2, 4) is y=-6x+16
3 0
3 years ago
I will give brainly help.
Burka [1]

Answer:

Company G charges 5 more dollars per hour than Company H.

5 0
2 years ago
Read 2 more answers
Please please helpp i will mark brainliest!!! 100 points !! ill mark brainliest if its correct.
Paraphin [41]

Answers and Step-by-step explanations:

1. We need to plug in values of x into the functions f(x) and g(x):

f(2) = 2.5 * 2 - 10.5 = -5.5                         g(2) = 64 * (0.5^2) = 16

f(3) = 2.5 * 3 - 10.5 = -3                            g(3) = 64 * (0.5^3) = 8

f(4) = 2.5 * 4 - 10.5 = -0.5                         g(4) = 64 * (0.5^4) = 4

f(5) = 2.5 * 5 - 10.5 = 2                             g(5) = 64 * (0.5^5) = 2

f(6) = 2.5 * 6 - 10.5 = 4.5                          g(6) = 64 * (0.5^6) = 1

2.

(a) Notice that at 14 on the x-axis, the blue line is above 4. Meanwhile, at 14, the red line is less than 2. Obviously, we can see that if the printed ad revenue was less than 2 and the online ad revenue was greater than 4, that's more than a double, so the lead marketing executive is correct.

(b) The approximate year that the revenues were equal is the place where the lines intersect. It's approximately at Year 7.5.

3.

(a) The Western Beach is clearly decreasing; it is decreasing by 2 ft per year. Meanwhile, the Dunes Beach is clearly increasing; it's increasing by 5 ft per year. (I'm not sure what kinds of "patterns" you're looking for).

(b) Year 11 and 12; this is because here, Western is going from 78 to 76 ft and Dunes is going from 75 to 80, so they'll likely coincide sometime in that time interval.

(c) One thing you can do is to create equations (of the form y = mx + b) for each of the beaches and then set those linear equations equal to each other.

6 0
3 years ago
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