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3 years ago

Write a verbal sentence to represent 3n-35=79

Mathematics

1 answer:

pantera1 [17]3 years ago

3 0

Answer: three n minus thirty-five equals seventy-nine

Step-by-step explanation:

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In a bag there are pink buttons, yellow buttons and blue buttons.

larisa86 [58]

Answer:

do you have any options for this question just to check??

8 0

2 years ago

Answers: <br> a) 74<br> b) 108<br> c)49<br> d)51

lisabon 2012 [21]

Answer:

Step-by-step explanation:

sum of exterior angles = 360

64 + 60 + 69 + 42 + x + x +23 = 360

2x + 258 = 360

2x = 360 - 258

2x = 102

x = 51

5 0

3 years ago

Read 2 more answers

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

Your school is hosting a play and tickets are being sold for $10 each the cost to rent the nearby theater and costumes is $450 a

svlad2 [7]

Answer:

Step-by-step explanation:

450 < 10x - 1x

450 < 9x

50 < x

you must sell 50 tickets to break even; 51 to make a profit

8 0

3 years ago

Can someone help me please

exis [7]

Answer:

43.7 ft

Step-by-step explanation:

Radius of the semicircular garden is 8.5 ft .

Diameter = 8.5 × 2 = 17 ft

Circumference of the garden = $\frac{1}{2}$ × π × diameter = $\frac{1}{2}$ × π × 17 = 26.7 ft (rounded up to the nearest tenth of a foot)

Distance around the gardent = Circumference + diameter = 26.7 ft + 17 ft = 43.7 ft

6 0

3 years ago