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3 years ago

Please help me with this question!!

Mathematics

1 answer:

klio [65]3 years ago

4 0

Answer:

Step-by-step explanation:

9/50 // 12 / 1 You have a 4 tier fraction here. You must invert the denominator (12/1) and multiply

9/50 * 1/12 Cancel 3 in the 9 and 3 in the 12

3/50 * 1 / 4 Multiply the numerators together and the denominators together.

3 / 200 each bottle with have 3/200 L or 1.5 / 100 L syrup in it.

1.5 / 100 L = 1.5 * 10 / 100 * 10 = 15 mL of medication in each bottle.

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8.14÷20 please show work

saul85 [17]

Answer:

0.407

Step-by-step explanation:

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3 years ago

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What is the value of

MakcuM [25]

Answer:

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3 years ago

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Find the probability that x <= -3

Slav-nsk [51]

Answer:

30%

Step-by-step explanation:

there are two numbers that fit in the x<=3 category, those being -3 and -5, so you add the two decimals for them (.17 and .13) to get .30 and move the decimal two times to the right to make it in percentage form.

6 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

a. Are the following triangles congruent? If so, which postulate or theorem proves this? b. What transformation moved one triang

xenn [34]

Answer:

Yes by SAS

Step-by-step explanation:

If you look at the triangles they have 2 side lenght in common (and since it is a right triange they have all the sides in commmon) and they share an angel

In other words 2 sides are conurent with an angle betwene them.

You would need to rotait the shape 90 degrese.

4 0

3 years ago