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3 years ago

Evaluate the expression (- n) ^ (n + n) + n ^ (n - n) , when n = 2

Mathematics

1 answer:

sladkih [1.3K]3 years ago

5 0

Answer:

Step-by-step explanation:

Substitute the value of the variable and simplify.

$(-(2))^{(2+2)} +2^{(2-2)}$

$-2x^{2+2} =2^{4} = 16$

$2^{2-2} =2^{0}$

Anything raised to the power of 0 is one.

16 + 1 = 17

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Homer is giving some cookies to each of

Stells [14]

Answer:

7 cookies

Step-by-step explanation:

We have 3 brothers

First , Second, Third

Let the total number cookies be represented by A

1 cookie = 1

Working backwards, we start from the third brother

If we work backwards

It means he gave everything away

We start from the youngest

He was given 1/2 of what was left and 1/2 a cookie

This means,

1/2 + 1/2 = 1

The second brother

He got half of what is left and 1/2 a cookies

Half of what is left from brother

= What the youngest brother got + 1/2 + 1/2 a cookie

= 1 + 1/2 + 1/2

= 1.5 + 1/2

= 2 cookies

For the first brother

He got 1/2 of the cookies + 1/2 cookie

= 1.5 × 2 + 1/2 + 1/2 cookies

= 3 1/2 + 1/2

= 4 cookies

The first brother got 4 cookies

The second brother got 2 cookies

The third broth got 1 cookies

3 0

3 years ago

It Takes Mike 18 minutes to finish reading 4 pages of a book. How long does it take him to finisn reading 30 pages?

Inga [223]

Answer:

135

Step-by-step explanation:

First, divide.

18 ÷ 4 = 4.5

Then, multiply.

4.5 × 30 = 135

Have a great day!

4 0

3 years ago

The volume of a cone is 329.6 cubic inches, and the height is 5.4 inches. Which of the following is the closest to the radius r

Anna [14]

Answer:

329.6=1/3×16.97r

5.66r=329.6/÷5.66

r=58.23

8 0

3 years ago

Where is the x-axis (where y is zero) in the inequality 5x+3y>450

valina [46]

X intercept:
let y=0
5x+3(0)>450
5x>450
450/5
x>90
y intercept:
let x=0
5(0)+3y>450
3y>450
450/3
y>150

3 0

4 years ago

Lim n-> infinity [1/3 + 1/3² + 1/3³ + . . . .+ 1/3ⁿ]

Verizon [17]

Answer:

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

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{n \to \infty }\rm \bigg[\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } \bigg]$

Let we first evaluate

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

Its a Geometric progression with

$\rm :\longmapsto\:a = \dfrac{1}{3}$

$\rm :\longmapsto\:r = \dfrac{1}{3}$

$\rm :\longmapsto\:n = n$

So, Sum of n terms of GP series is

$\rm :\longmapsto\:S_n = \dfrac{a(1 - {r}^{n} )}{1 - r}$

$\rm :\longmapsto\:S_n = \dfrac{1}{3} \bigg[\dfrac{1 - {\bigg[\dfrac{1}{3} \bigg]}^{n} }{1 - \dfrac{1}{3} } \bigg]$

$\rm :\longmapsto\:S_n = \dfrac{1}{3} \bigg[\dfrac{1 - {\bigg[\dfrac{1}{3} \bigg]}^{n} }{\dfrac{3 - 1}{3} } \bigg]$

$\rm :\longmapsto\:S_n = \dfrac{1}{3} \bigg[\dfrac{1 - {\bigg[\dfrac{1}{3} \bigg]}^{n} }{\dfrac{2}{3} } \bigg]$

$\bf\implies \:S_n = \dfrac{1}{2}\bigg[1 - \dfrac{1}{ {3}^{n} } \bigg]$

<u>Hence, </u>

$\bf :\longmapsto\:\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } = \dfrac{1}{2}\bigg[1 - \dfrac{1}{ {3}^{n} } \bigg]$

<u>Therefore, </u>

$\purple{\rm :\longmapsto\:\displaystyle\lim_{n \to \infty }\rm \bigg[\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } \bigg]}$

$\rm \: = \: \displaystyle\lim_{n \to \infty }\rm \dfrac{1}{2}\bigg[1 - \dfrac{1}{ {3}^{n} } \bigg]$

$\rm \: = \: \rm \dfrac{1}{2}\bigg[1 - 0 \bigg]$

$\rm \: = \: \rm \dfrac{1}{2}$

<u>Hence, </u>

$\purple{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{n \to \infty }\rm \bigg[\dfrac{1}{3} + \dfrac{1}{ {3}^{2} } + \dfrac{1}{ {3}^{3} } + - - + \dfrac{1}{ {3}^{n} } \bigg]} = \frac{1}{2}}}$

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<h3><u>Explore More</u></h3>

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1}}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} = 1}}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} = 1}}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} = 1}}$

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} = loga}}$

8 0

3 years ago