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Alex73 [517]
3 years ago
15

The volume of a rectangular prism can be found by multiplying the three side lengths. A bookcase in the shape of a rectangular p

rism has dimensions (b+8) ft, b ft, and (b+3) ft. Find the volume of the bookcase​
Mathematics
1 answer:
Vesna [10]3 years ago
3 0

Answer:

(b³+11b²+24)ft³

Step-by-step explanation:

Volume of the rectangular prism = Length * Width * Height

Given the dimension

Length = b+8 ft

Width = b ft

Height = (b+3)ft

Volume of the rectangular prism = b(b+8)(b+3)

Volume of the rectangular prism = b(b^2+3b+8b+24)

Volume of the rectangular prism = b(b^2+11b+24)

Volume of the rectangular prism = b(b^2)+b(11b)+1(24)

Volume of the rectangular prism = b^3+11b^2+24

Hence the Volume of the rectangular prism is (b³+11b²+24)ft³

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B

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2 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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3 years ago
It take Rick 5 minutes to wash 4 dishes. If he continue to work at this rate.how many minutes will it take Rick to wash 24 dishe
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5 minutes         x minutes

---------------  =   -------------

4 dishes           24 dishes


using cross products

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divide each side by 4

5*24/4 = x

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Today, everything at a store is on sale. The store offers a 15% discount. The regular price of a T-shirt is $22. What is the dis
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The discount price is $18.7

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Answer:

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And BC= 36

36-14.4= y

21.6= y= DC

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