Which Figure is a rotation of the Blue Figure? A B C D

2. Which fraction is in its simplest form? 2/4 2/8 2/6 2/5

wariber [46]

Answer:

2/5

Step-by-step explanation:

2 / 4 could be divided by two in both sides.

2 ÷ 2 = 1

4 ÷ 2 = 2

2/4 = 1/2

Same goes for 2/8 and 2/6

2 ÷ 2 = 1

8 ÷ 2 = 4

2/8 = 1/4

2 ÷ 2 = 1

6 ÷ 2 = 3

2/6 = 1/3

2/5 however, could not be divided by two on both sides because there is 5 and it is not a factor of 2. So it is in its simplest form

7 0

3 years ago

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HELP WITH THIS EZ QUESTION

9966 [12]

Answer:

89

Step-by-step explanation:

$3+75t-16t^{2} =h\\\\$

we evaluate when t=2

$3+75(2)-16(2)^2=h\\\\3+150-64=h\\\\h=89$

so it's 89 feet

7 0

3 years ago

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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

Find the geometric sum: 40 + 20 + 10 + … + 0.078125

pentagon [3]

This algebra lesson explains geometric series. ... To find the sum of the first n terms of a geometric sequence: Sn = the summation of ... 5 , -10 , 20 , -40 , 80 , .

6 0

3 years ago

Brainliest, 50 points

Alex73 [517]

10y+50

Answer:

10(y+5)=10y+50

just

multiplying both by 10.Use distribute property.

5 0

3 years ago

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