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Stels [109]
3 years ago
10

Write (2x10) + (9 x 1) + (7 x 0.10 ) + (8 x 0.001) in standard form.

Mathematics
1 answer:
bonufazy [111]3 years ago
3 0

Answer:

hi

Step-by-step explanation:

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The ratio of roses to carnations is 7 to 5. If 96 flowers are needed for table decorations,
coldgirl [10]

Answer: 56 roses and 40 carnations are needed.

Step-by-step explanation: <u>Here’s the table of ratios that’ll help you find how many roses and carnations are needed to make a total of 96 flowers for table decorations:</u>

7:5 /14:10 /21:15 /28:20 /35:25 /42:30 /49:35/ 56:40/

*56 + 40 = 96

Hence, 56 roses and 40 carnations are needed if 96 flowers are needed for table decorations.

<u></u>

8 0
2 years ago
Someone please help me will give BRAILIEST!!!!!!
rewona [7]
The answer is true plz thnk me
4 0
3 years ago
Read 2 more answers
Calculate the total cost of an item bought at a negotated price of $13,290 with 5.5 percent sales tax and a $125 registration fe
Charra [1.4K]
C=13290+13290(.055)+125=$14145.95<span>
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3 years ago
Name the maximum and minimum of the function
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The alimentary canal and the <span>gastrointestinal tract (GI tract)</span>
8 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
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