All categories

2 years ago

Which is the perimeter for the following?

Mathematics

1 answer:

ElenaW [278]2 years ago

8 0

Step-by-step explanation:

Add thel and multiply the 2

You might be interested in

What is the formula for distance?

SOVA2 [1]

Answer:

Distance formula is below

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Right Triangle with legs of 15 and 20, what is the length of the Hypotenuse?

ioda

15^2 + 20^2
= 225 + 400
= 625
square root of 625 = 25

answer:<span>length of the Hypotenuse is</span> 25

5 0

3 years ago

After solving -4x -12 = 4x + 12, you get 8x = 0, what is the answer?

Paha777 [63]

The answer of this is x=0

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

Determine the area of the figure

aliya0001 [1]

The area of the trapezoid is x+11 cm.

Given a trapezoid has one side of (x+6) cm, the other side is 5 cm, and the height is 2 cm.

A two-dimensional quadrilateral consisting of the sum of non-adjacent parallel sides and a suitable non-parallel side is denoted as a trapezoid.

Now we will use the formula to find the area of a trapezoid

Area = 1/2 × (base 1 + base 2) × height

Here base 1 = (x+6) cm, base 2 = 5cm and height = 2cm

Substituting the values into the formula, we get

Area = 1/2 × (x+6+5) × 2

Area = 1/2 × (x+11) × 2

Area = x+11

Thus, the area of the given figure when one side is x+6, the other side is 5 cm and the height is 2 cm is x+11 cm.

Learn more about the area of trapezium from here brainly.com/question/15815316

#SPJ1

8 0

2 years ago