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2 years ago

Could someone please explain surface area to me?Thanks will be rewarded

2 answers:

7 0

Surface area of a solid is the sum of all areas of the faces of the shape so if you have the net set up you will just have to find the area of each face and add them up!

solong [7]2 years ago

5 0

Ok its like for rectangular prisim its the volume here is a formula V=lwh. There you go use that for anything i do k12 so message me if you need anything.

You might be interested in

Find the value of x in the proportion. Check your answer. 35/x=7/13 x=?

maks197457 [2]

This is super simple. Begin with cross multiplying

35/x=7/13 becomes (35)*(13)=(7) *(x)

Then multiply my love muffin, you get...

455=7x

divide both sides by 7

65=x
-or-
x=65

hope that helped :)

6 0

2 years ago

What is the answer to 3x+2x+7=22

gavmur [86]

Answer:

Step-by-step explanation:

3x+2x+7=22

5x+7-7=22-7

5x=15

5x/5=15/5

X=3

8 0

2 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

2 years ago

Estimate. 40,182 : 2 Choose 1 answer: А 20 B 200 2,000 20,000

Anvisha [2.4K]

Answer:

20,000

Step-by-step explanation:

3 0

2 years ago

Two lines intersect in more than one point.

Paraphin [41]

So, either x or y axis is the same for that intersect point