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3 years ago

What is the recursive rule for an=4n−1?

Mathematics

2 answers:

Zolol [24]3 years ago

8 0

The correct answer for the first one is: <span>a1=3;an=<span>an−1</span>+4 </span>

tensa zangetsu [6.8K]3 years ago

7 0

This is the answer, i have completed the test.

Hope this helps someone! Good luck :)

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LOTS OF POINTS, WILL MARK BRAINLIEST! pls show working!!!!!

sp2606 [1]

<em>Answer:</em> ΔTAU ≈ ΔUAV ≈ ΔTUV

<em>Step-by-step explanation:</em>

I'm not really sure what "work" you really need; this is a problem that can be solved easily by simply looking at the triangles and seeing which sides have the same ratio of distances for each side.

Best of luck with your assignment. :) Feel free to give me Brainliest if you feel this helped. Have a good day.

8 0

2 years ago

A rectangle has a width of 46 centimeters and a perimeter of 204 centimeters. What is the rectangle’s length?

Vanyuwa [196]

Answer:

Length = 56

Step-by-step explanation:

Perimeter = Length + Length + Width + Width

Perimeter = 2L + 2W

2L + 2W = 204

2L + 2(46) = 204

2L + 92 = 204

Subtract 92 from both sides

2L = 112

Divide both sides by 2

L = 56

Length = 56

7 0

4 years ago

Rick took a 20-minute coffee break. at the end of the break, it was 10:00

gladu [14]

Rick started his coffee break at 9:40 am.

4 0

3 years ago

Read 2 more answers

Lim n→∞[(n + n² + n³ + .... nⁿ)/(1ⁿ + 2ⁿ + 3ⁿ +....nⁿ)]

Schach [20]

Step-by-step explanation:

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

To, evaluate this limit, let we simplify numerator and denominator individually.

So, Consider Numerator

$\rm :\longmapsto\:n + {n}^{2} + {n}^{3} + - - - + {n}^{n}$

Clearly, if forms a Geometric progression with first term n and common ratio n respectively.

So, using Sum of n terms of GP, we get

$\rm \: = \: \dfrac{n( {n}^{n} - 1)}{n - 1}$

$\rm \: = \: \dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }$

Now, Consider Denominator, we have

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {n}^{n}$

can be rewritten as

$\rm :\longmapsto\: {1}^{n} + {2}^{n} + {3}^{n} + - - - + {(n - 1)}^{n} + {n}^{n}$

$\rm \: = \: {n}^{n}\bigg[1 +\bigg[{\dfrac{n - 1}{n}\bigg]}^{n} + \bigg[{\dfrac{n - 2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]$

$\rm \: = \: {n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]$

Now, Consider

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} }$

So, on substituting the values evaluated above, we get

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\dfrac{ {n}^{n} - 1}{1 - \dfrac{1}{n} }}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n} - 1}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{ {n}^{n}\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty} \frac{1}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} + - - - + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}$

Now, we know that,

$\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to \infty} \bigg[1 + \dfrac{k}{x} \bigg]^{x} = {e}^{k}}}}$

So, using this, we get

$\rm \: = \: \dfrac{1}{1 + {e}^{ - 1} + {e}^{ - 2} + - - - - \infty }$

Now, in denominator, its an infinite GP series with common ratio 1/e ( < 1 ) and first term 1, so using sum to infinite GP series, we have

$\rm \: = \: \dfrac{1}{\dfrac{1}{1 - \dfrac{1}{e} } }$

$\rm \: = \: \dfrac{1}{\dfrac{1}{ \dfrac{e - 1}{e} } }$

$\rm \: = \: \dfrac{1}{\dfrac{e}{e - 1} }$

$\rm \: = \: \dfrac{e - 1}{e}$

$\rm \: = \: 1 - \dfrac{1}{e}$

Hence,

$\boxed{\tt{ \displaystyle\lim_{n \to \infty} \frac{n + {n}^{2} + {n}^{3} + - - + {n}^{n} }{ {1}^{n} + {2}^{n} + {3}^{n} + - - + {n}^{n} } = \frac{e - 1}{e} = 1 - \frac{1}{e}}}$

3 0

3 years ago

Polygon ABCD is plotted on a coordinate plane and then rotated 90° clockwise about point C to form polygon A′B′C′D′. Match each

horrorfan [7]

The vertices of ABCD after 90 degrees clockwise rotation about point C are: A' (0,6), B' (3,7), C' (4,6) and D' (4,3)

<h3>How to match the vertices of the polygon?</h3>

The image of the polygon ABCD is not given; however, the question can still be answered because the coordinates are known.

The vertices of polygon ABCD are given as:

A = (4, 6)

B = (5, 3)

C = (4, 2)

D = (1, 2)

The rule of rotation about point C is:

(x,y) = (a + b - y, x + b - a)

Where:

(a, b) = (4, 2) --- the point of rotation.

So, we have:

(x,y) = (4 + 2 - y, x + 4 - 2)

(x,y) = (6 - y, x + 2)

The above means that:

A' = (6 - 6, 4 + 2) = (0,6)

B' = (6 - 3, 5 + 2) = (3,7)

C' = (6 - 2, 4 + 2) = (4,6)

D' = (6 - 2, 1 + 2) = (4,3)

Hence, the image of the rotation and their vertices (i.e. coordinates) are:

A' = (0,6)

B' = (3,7)

C' = (4,6)

D' = (4,3)