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3 years ago

What is the SCALE FACTOR (in decimal form) for a ratio of 6:15?

Mathematics

2 answers:

deff fn [24]3 years ago

5 0

0.4
.....

......
......

NikAS [45]3 years ago

5 0

Answer:

0.4

Step-by-step explanation:

You need to divide the 6 by 15 in order to get 0.4.

Ratios, fractions, decimals, and percentages are all pretty much each other, as ratios are just fractions but sideways with the colon in between the numbers. Ratios are to show a relationship, as well as the others. Fractions and ratios are decimals, as you can just simplify them into decimals. Percentages are just decimals but with the decimal moved over to the left twice. So 0.43 becomes 43%.

1.49 becomes 149%

0.00432 becomes 0.432%

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What is 56/18 in simplest form

Triss [41]

Answer:

The answer in the simplest form would be 28/9 or 3 1/9

Step-by-step explanation:

Very simple you would divide the 56 and the 18 by 2 and you get 28/9.

or you could change it to a improper fraction in that case it would be 3 1/9

6 0

3 years ago

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Grace, Chelsea, and Roan are simplifying the same polynomial expression. which students work is correct and why?

Trava [24]

Answer:

Grace

Step-by-step explanation:

Chelsea and Roan didn't distribute the negative sign in the second half of the first expression.

-2(x - 8) is

-2(x) + (-2)(-8)

-2x + 16

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3 years ago

Read 2 more answers

If anyone can check my work for me that would be greatly appreciated I'm having trouble.

krek1111 [17]

The answer is -13. Solution: = |-4b - 8| + |-1 - b^2| + 2b^3 = |-4(-2) - 8| + |-1 - -2^2| + 2(-2)^3 = |8-8| + |-1+4| + 2(-8) = |0| + |3| + (-16) = -13

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

Help me , i need fast

MAVERICK [17]

I have a feeling it B but I don’t know maybe C

7 0

3 years ago