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3 years ago

Which numbers are polygons

Mathematics

2 answers:

OlgaM077 [116]3 years ago

6 0

c and a

by the way, thats easy. NO OFFENCE XD

Lynna [10]3 years ago

3 0

C and a !!!!!! Jkjkggynfyjfddsd

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Geometry matching help

o-na [289]

Answer:

Ellipse C

Circle E

Sphere B

Great Circle D

Triangle A

Step-by-step explanation:

Cross sections are slices through a 3d figure to create a 2d shape. A cone creates a circle and triangle. A sphere creates a great circle. A cylinder creates an ellipse. Rotating a semi circle around an axis creates a sphere.

7 0

3 years ago

Five times the sum of a number and 3 is the same as 3 multiplied by 1 less than twice the number

Aneli [31]

Answer:-2

Step-by-step explanation:

(n + 3) 5 = 3 x 1 - n

distribute so 5n + 15= 3 - n

subtract three from both sides

5n - 12 = -n

subtract 5n from both sides

12 = −6n

Divide both sides by -6

n=-2

7 0

3 years ago

Bananas cost $ 0.59 per pound.Write an equation that could be used to find the total cost,y,of x pounds of bananas.

maks197457 [2]

I think it is 5 Bc I do and if you got to know so bad

4 0

3 years ago

Read 2 more answers

(2,-4) is reflected across the x-axis. what are the coordinates of its image?

masha68 [24]

Answer:

(2, 4)

Step-by-step explanation:

When you reflect/flip over an axis, negate value of the opposite of it. Fox the X axis, negate the y value.

5 0

2 years ago

Read 2 more answers

Lim (n/3n-1)^(n-1) n → ∞

n200080 [17]

Looks like the given limit is

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}$

With some simple algebra, we can rewrite

$\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)$

then distribute the limit over the product,

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}$

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, e :

$\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n$

To make our limit resemble this one more closely, make a substitution; replace 9/(n - 9) with 1/m, so that

$\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1$

From the relation 9m = n - 9, we see that m also approaches infinity as n approaches infinity. So, the second limit is rewritten as

$\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}$

Now we apply some more properties of multiplication and limits:

$\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9$

So, the overall limit is indeed 0:

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}$

7 0

3 years ago