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3 years ago

10

A store that buys and sells used video games buys a game at a cost of $29 and sells it at a selling price of $33. Find the mar

kup. Then find the percent markup.

1 answer:

Rudiy273 years ago

3 0

$4. Not sure about the percent though, sorry!

Hope it helped :)

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List the steps that should be used to solve the equation f(x) = g(x) by graphing when f(x) = (1/2)² - 3 and g(x) = 3x + 5?

jolli1 [7]

value of x is: $x=\frac{-31}{12}$

Step-by-step explanation:

We need to solve the equation f(x)=g(x)

where $f(x)=(\frac{1}{2})^2-3$

and $g(x) = 3x + 5$

Solving to find f(x)=g(x)

Putting values of f(x) and g(x)

$f(x)=g(x)\\(\frac{1}{2})^2-3=3x + 5\\\frac{1}{4}-3=3x + 5\\Subtarct\,\,5\,\,on\,\,both\,\,sides:\\\frac{1}{4}-3-5=3x + 5-5\\\frac{1}{4}-8=3x\\Taking\,\,LCm\,\,on\,\,left\,\,side\\\frac{1-8*4}{4}=3x\\\frac{-31}{4}=3x\\Divide\,\,both\,\,sides\,\,by\,\,3\\x=\frac{-31}{4*3}\\x=\frac{-31}{12}$

So, value of x is: $x=\frac{-31}{12}$

The graph will be a straight line x= -2.58 or x = -31/12

Keywords: Composition of Functions

Learn more about Composition of Functions at:

#learnwithBrainly

7 0

3 years ago

Determine whether the sequences converge.

Alik [6]

$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

and so on. We're getting an inkling that the explicit closed form for the sequence may be $a_n=2^{(2^n-1)/2^n}$ , but that's not what's asked for here. At any rate, it appears reasonable that the exponent will steadily approach 1. Let's prove this.

Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

$a_{k+1}=\sqrt{2a_k}$

and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

4 0

3 years ago

Please help me. Dont judge.

tia_tia [17]

It would be A because u need to add 5

6 0

2 years ago

Read 2 more answers

I will give brainliest

Rom4ik [11]

Answer:

plz give it to me even though i didn't anwser it I only got brianlyist onece

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Complete the table to show the interest earned for different savings principals, interest rates, time periods.

EastWind [94]

Answer:just multiply your numbers by the interest rates with a decimal point in front.

Step-by-step explanation:

7 0

2 years ago