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2 years ago

7

The stores below each have the same snowboard for an original price of $189.59. At which store can you get the snowboard for the

lowest price? Store A: Sale of 30% off and a successive discount of 10% off. Store B: Sale of 20% off and a successive discount of 20% off. Store C: Sale of 40% off. a. All of the stores have the same discounted price. b. You can get the lowest price at Store A. c. You can get the lowest price at StoreB. d. You can get the lowest price at Store C.

1 answer:

Rama09 [41]2 years ago

7 0

Answer: D.

You can get the lowest price at Store C

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Answer:

x=60.9°

Step-by-step explanation:

Given that the height of ball from the ground is 150ft

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Trey's horizontal line of sight is 6 feet above ground, then;

The height of ball from Trey's horizontal line of sight is;

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To find the angle x, assume a triangle with a base of 80 ft , a height of 144 ft and a slant height that represent the line of sight at an angle x

To get angle x , you apply the tangent of an angle formula where;

tan Ф°= length of opposite site of the angle/length of the adjacent side of the angle

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3 years ago

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Answer:

Add 4 to both sides

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Step-by-step explanation:

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3 years ago

A store uses the expression –2p + 50 to model the number of backpacks it sells per day, where the price, p, can be anywhere from

Grace [21]

Your question is store uses the expression –2p + 50 to model the number of backpacks it sells per day, where the price, p, can be anywhere from $9 to $15. Which price gives the store the maximum amount of revenue, and what is the maximum revenue?

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2 years ago

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Which whole number is closest to 577.41 closest to?

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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).

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