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2 years ago

The figures are similar. Find each missing measure.

Mathematics

2 answers:

Elena L [17]2 years ago

7 0

Answer:

Given that, figure LMNO and figure PQRS are similar

So, RQ / MN = SP / OL

x km / 6 km = 9 km / 27 km
x km / 6 km = 1 km / 3 km
x km = 6 km / 3 km
<u>x = 2 km</u>
<u>b) 2</u> is the right answer.

zmey [24]2 years ago

4 0

Find the ratio:

9/27 = 1/3

6 x 1/3 = 2

Answer: b. 2

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The school that Kathryn goes to is selling tickets to the annual talent show. On the first day of ticket sales the school sold 1

Kobotan [32]

Answer:

Step-by-step explanation:

a+s=14

a=14-s

a+8s=56, using a from above in this equation gives

14-s+8s=56

14+7s=56

7s=42

s=6

a=14-s

a=14-6=8

So student tickets cost $6 and adult tickets cost $8.

5 0

2 years ago

. For the function given, state the starting point for a sample period:

Aneli [31]

I think the answer is π?

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

How many solutions for x does the following equation have?

Zepler [3.9K]

Answer:

c. 0

Step-by-step explanation:

3(x+8) - 1 = 3x+4

3x+24-1 = 3x+4

3x+23 = 3x+4

There are no solutions since both equations have the same slope meaning they are parallel and they will never intersect since they have different y-intercepts.

6 0

1 year ago

You helped organize a birthday party for twenty kids. You filled 150 water balloons and the kids smashed 4/5 of them in ten minu

notka56 [123]

Answer:

Step-by-step explanation:

Okay, to find the number of balloons left over, disregard the number of kids and focus only on 150 balloons and that only 4/5 were "smashed." Multiply 150 by 4/5 to find the amount of balloons used, which is 120, and find the difference between 150 and 120, which is 30.

4 0

3 years ago