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madam [21]
2 years ago
14

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

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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
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