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3 years ago

A two-dimensional figure with four sides has 2 pairs of opposite parallel sides, 4 congruent sides, and 4 right angles.

Mathematics

2 answers:

kap26 [50]3 years ago

8 0

Parallelogram rectangle square rhombus

sashaice [31]3 years ago

7 0

Answer:

Parallelogram, rectangle, square and rhombus

Step-by-step explanation:

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lyudmila [28]

I believe it is 5 or 10

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3 years ago

(2x-3)(x+5) SHOW WORK FOR POINTS

vagabundo [1.1K]

(2x-3)(x+5)
Use foil to simplify, First multiply together the first terms of the pairs, then the outer terms, the middle terms, and finally the last terms. Then combine like terms.
2x^2+10x-3x-15 equals 2x^2+7x-15

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3 years ago

If answer is correct I will mark as brainliest.

Ilia_Sergeevich [38]

Answer:

No positive real solutions.So the answer is zero.

Step-by-step explanation:

8 0

3 years ago

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

What is the value of x? 3x + (6x - 6) = 4(2x-2)

olya-2409 [2.1K]

Answer:

x = -2

Step-by-step explanation:

To start, let's add on the first side to get 3x + 6x - 6 = 9x - 6. For the second side, we need to distribute the 4 to get 4(2x) + 4(-2), or 8x + (-8) or 8x - 8. Setting these equal, we subtract 8x from both sides to get x - 6 = -8, and adding 6 to both sides gives x = -2.

8 0

3 years ago