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Soloha48 [4]
3 years ago
5

Order the temperatures for the week from hottest to coldest: 25 degrees, 2 degrees below zero, -16 degrees, and 40 degrees above

zero.
A. -2°, -16°, 25°, 40°
B. -16°, -2°, 259, 16°
C. 40°, 25°, -16°, -2°
D. 40°,25°, 2°, -16°
E. 40°, 250, -2°, -16°
Mathematics
1 answer:
Strike441 [17]3 years ago
3 0

Answer:

C. 40,25,-16,-2 Celsius

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Find the volume of a prism with base area 32 cm2 and height 1.5 cm.
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Answer: 48

Step-by-step explanation:

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2 years ago
I WILL GIVE BRAINLIST AND 5 STARS <br> Show that (2x + i)(2x – i) = 4x² + 1
sladkih [1.3K]

Answer:

see below

Step-by-step explanation:

(2x + i)(2x – i)

FOIL

first  2x*2x =4x^2

outer  -i * 2x = 2x i

inner i * 2x = 2x i

last i * -i = -i^2  and i^2 = -1  so - (i^2 ) = - (-1) = 1

Add together

4x^2 + 2xi -2x i +1

Combine like terms

4x^2 +1

8 0
3 years ago
A student take out a simple interest loan for $2000.00 for 6 month at a rate of 6.50%. Find the interest the student must pay
ikadub [295]

Answer: he must pay $650 interest

Step-by-step explanation:

2000 x 0.650 x 1/2 = 650

5 0
3 years ago
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
Which of the tables represents a function? Table P Input Output 8 3 1 7 5 4 Table Q Input Output 9 3 9 5 4 2 Table R Input Outpu
Colt1911 [192]

Answer:

Table P represents a function

Step-by-step explanation:

* Lets explain the meaning of the function

- A function is a relation between a set of inputs and a set of outputs

 in condition of each input has exactly one output

- Ex:

# The relation {(1 , 2) , (-4 , 5) , (-1 , 5)} is a function because each x in the

   order pair has only one value of y

# The relation {(1 , 2) , (1 , 5) , (3 , 7)} is not a function because there is x

   in the order pairs has two values of y (x = 1 has y = 2 and y = 5)

* Lets solve the problem

# Table P :

- In put    : 8  ,  1  ,  5

- Out put : 3  ,  7  , 4

∵ Each input has only one output

∴ Table P represents a function

# Table Q :

- Input     : 9  ,  9  ,  4

- Out put : 3  ,  5  ,  2

∵ The input 9 has two outputs 5 and 2

∴ Table Q doesn't represent a function

# Table R :

- In put    : 7  ,  8  ,  7

- Out put : 2  ,  6  ,  3

∵ The input 7 has two outputs 2 and 3

∴ Table R doesn't represent a function

# Table S :

- In put    :  1  ,  1  ,  9

- Out put :  7 ,  5  ,  2

∵ The input 1 has two outputs 7 and 5

∴ Table S doesn't represent a function

* Table P represents a function

4 0
4 years ago
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