All categories

3 years ago

Pick all the names for this shape. Sides that appear to be parallel are parallel

Mathematics

2 answers:

Lesechka [4]3 years ago

4 0

The correct answer is A

Wewaii [24]3 years ago

3 0

Answer:

I would say a parrallelogram

Step-by-step explanation:

You might be interested in

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

Plssssss help giving brainliest

ioda

Answer:

ask your teacher

Step-by-step explanation:

3 0

3 years ago

Vector B has x, y, and z components of 3, 7.2,

Temka [501]

Answer:

$\approx 9.0\:\mathrm{units}$

Step-by-step explanation:

The magnitude of 3D vector $\vec{v}$ can be given by the following:

$| \overline{v}| =\sqrt{{v_x}^2+{v_y}^2+{v_z}^2}$

Plugging in given values, we have:

$| \overline{v}|=\sqrt{3^2+7.2^2+4.5^2},\\| \overline{v}| \approx \fbox{$9.0\:\mathrm{units}$}$ .

6 0

2 years ago

the ratio of students with bikes to students with scooters in the school is 85:51. What is the greatest common factor that you c

Vikentia [17]

Answer:GCF=17

To find the greatest common factor, or GCF, of the numbers, we first need to list the prime factors of both 85 and 51

52=3*17

85=5*17

17 is the sole common factor that these two numbers have in common, therefore 17 is the GCF.

To simplify the ratio, simply divide both numbers by 17.

85:51 becomes 5:3

5 0

3 years ago

Read 2 more answers

PLS HELP ME ITS DUE IN 10 MIN------------- HERE ARE THE ASNWERS

DENIUS [597]

5x + 20 because rise over run 5 over 1 so its 5x and 20 is on the y intercept line

Step-by-step explanation:

4 0

2 years ago