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3 years ago

A 54-kg jogger is running at a rate of 3 m/s. What is the kinetic energy of the jogger? A. 18 J B. 81 J C. 162 J D. 243 J

Mathematics

2 answers:

Mrrafil [7]3 years ago

5 0

Answer:

The answer would be be D.243 j

Step-by-step explanation:

Wewaii [24]3 years ago

4 0

$\displaystyle\bf\\\text{\bf kinetic energy}=\frac{mv^2}{2}=\frac{54\times3^2}{2}=27\times9=\boxed{\bf243~J} \\\\\text{\bf Correct answer: D.}$

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-8m+(-15)+4m-2r-10 ?

AVprozaik [17]

Answer:

- 4m - 2r - 25

Step-by-step explanation:

-8m + (-15) + 4m - 2r - 10

- 8m - 15 + 4m - 2r - 10

collect the like terms beginning with the positives

4m - 8m - 2r - 10 - 15

- 4m - 2r - 25

7 0

2 years ago

One positive integer is 5 times another positive integer and their product is 320. What are the positive integers?

seropon [69]

Answer:

The integers are (x, y) = (40, 8).

Step-by-step explanation:

x = 5y

xy = 320

Substitute the first equation into the second equation.

(5y)(y) = 320

5y^2 = 320

y^2 = 64

y = 8 (y must be positive)

The integers are (x, y) = (40, 8).

6 0

2 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

2 years ago

A local kids play sold 8 adult tickets and 9 child tickets for a total of $205. On day two they sold only 4 adult tickets and 3

Naya [18.7K]

Answer:

The price of 1 adult ticket is $ 11 and price of 1 child ticket is $13

Step-by-step explanation:

Let x be the cost of 1 adult ticket

Let y be the cost of 1 child ticket

Cost of 8 adult tickets = 8x

Cost of 9 child tickets = 9y

We are given that A local kids play sold 8 adult tickets and 9 child tickets for a total of $205

So, 8x+9y=205 ----- 1

Cost of 4 adult tickets = 4x

Cost of 3 child tickets = 3y

We are given that 4 adult tickets and 3 child tickets for a total of $83.

4x+3y=83 ----2

Plot 1 and 2 on graph

8x+9y=205 -- Red line

4x+3y=83 -- Blue line

Intersection point provides the solution

So, Intersection point =(11,13)

So, The price of 1 adult ticket is $ 11 and price of 1 child ticket is $13

4 0

3 years ago

How do i find the missing measurement?

lisabon 2012 [21]

What do u mean by missing measurement?

7 0

3 years ago