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jeka94
4 years ago
8

A smartphone originally worth $790 loses value at a rate of $175 each year.Write an equation to represent the value of the phone

, then find value of the phone after 4years.Identify your variables
Mathematics
1 answer:
7nadin3 [17]4 years ago
6 0
790-175x4=
take 175 and multiply it by 4 which gives you 700
Then subtract 700 from 790 and you get $90 after 4 years.
Your variables are 175 and 4
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mario62 [17]
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4 years ago
Over the course of a month, caught three times as many squirrels as deer. She also caught five less rabbits than squirrels. The
Westkost [7]

Answer:

The number of rabbits Katniss caught = 10 rabbits

Step-by-step explanation:

The number of squirrels Katniss caught = 3 × (The number of deer she caught)

The number of rabbits Katniss caught = The number of squirrels she caught - 5

The total number of animals she caught = 10 + 4 × (The number of deer she caught)

Let the number of deer Katniss caught = D

∴ The number of squirrels she caught = 3×D

The number of rabbits Katniss caught =  3×D - 5

The total number of animals she caught = 10 + 4 × D

Therefore, we have;

The number of squirrels she caught + The number of deer she caught + The number of rabbits she caught = 10 + 4 × D

3×D + D + 3×D-5 = 10 + 4 × D

7·D - 5 = 10 + 4·D

7·D - 4·D = 10 + 5

3·D = 15

D = 15/5 = 5

∴ The number of deer Katniss caught = D = 5

The number of rabbits Katniss caught =  3×5 - 5 = 10

The number of rabbits Katniss caught = 10 rabbits.

6 0
3 years ago
Fine the smallest positive integer n so that
Marizza181 [45]

Answer:

n = 9

Step-by-step explanation:

Let's first prove that for any constants k > 0, n\geq 1

\lim_{x\to \infty}\frac{\log x^k}{x^n}=0

The derivative

(\log (x^k))'=\frac{kx^{k-1}}{x^k}=\frac{k}{x}

and the derivative (x^n)' = nx^{n-1}

Now, applying L'Hôpital's rule we find that

\lim_{x\to \infty}\frac{\log x^k}{x}=\lim_{x\to \infty}\frac{k}{nx^n}=0

Now, let f be the function

f(x)=x^8log(x^3)+x^6log(x^5)

It is easy to see that  f(x) is O(x^n) only if n\geq 9

If n\geq 9  

\frac{f(x)}{x^n}=\frac{x^8log(x^3)+x^6log(x^5)}{x^n}=\frac{log(x^3)}{x^{n-8}}+\frac{log(x^5)}{x^{n-6}}

but both n-8 and n-6 are greater than one, so

\lim_{x \to \infty}\frac{f(x)}{x^n}=0

and f is O(x^n)

On the other hand, if n \leq 8 then  

\frac{f(x)}{x^n}=x^{8-n}log(x^3)+x^{6-n}log(x^5)

but 8-n is greater or equal than one, so

\lim_{x \to \infty}\frac{f(x)}{x^n}=\infty

and so f(x) in not O(x^n)

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