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3 years ago

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While on a train trip, Marsha noticed that there were 35 people in her train car. After the next stop, there were only 21 people

in her car. If p represents the percent decrease in the number of people in Marsha’s train car, what is p?

2 answers:

babunello [35]3 years ago

7 0

Answer:

the answer is 40

Step-by-step explanation:

because i big smart and because i did the answer:/

const2013 [10]3 years ago

5 0

Answer:

40%

Step-by-step explanation:

To find a percent, divide the part by the total and multiply by 100.

To find the part, subtract 21 from 35 to find the people who left.

35 - 21 = 14

(14/35)*100 = p

.4 *100 = p

40 =p

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consider the function and then use calculus to answer the questions that follow 1 1/x 5/x^2 1/x^3 (a) Find the interval(s) where

boyakko [2]

Answer:

a) $X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)$

b) $Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0 )$

Step-by-step explanation:

From the question we are told that

The Function

$f(x)=1+\frac{1}{x} +\frac{5}{x^2} +\frac{1}{x^3}$

Generally the differentiation of function f(x) is mathematically solved as

$f(x)=1+\frac{1}{x} +\frac{5}{x^2} +\frac{1}{x^3}$

$f(x)=\frac{x^3+x^2+5x+1}{x^2}$

Therefore

$f'(x)=\frac{x^2+10x+3}{x^4}$

Generally critical point is given as

$f'(x)=0$

$\frac{x^2+10x+3}{x^4}=0$

$x=-5 \pm\sqrt{22}$

Generally the maximum and minimum x value for critical point is mathematically solved as

$f'(-5 \pm\sqrt{22})$

Where

Maximum value of x

$f'(-5 +\sqrt{22})$

Minimum value of x

$f'(-5 +\sqrt{22})$

Therefore interval of increase is mathematically given by

$f'(-5 -\sqrt{22}),f'(-5 +\sqrt{22})$

$f(x)$

Therefore interval of decrease is mathematically given by

$(-\infty,-5 -\sqrt{22}),f'(-5 +\sqrt{22},0),(0,\infty)$

Generally the second differentiation of function f(x) is mathematically solved as

$f''(x)=\frac{2(x^2+15x+6)}{x^5}$

Generally the point of inflection is mathematically solved as

$f''(x)=0$

$x^2+15x+6=0$

Therefore inflection points is given as

$x=\frac{1}{2} (-15 \pm \sqrt{201}$

$f''(x)>0,\frac{1}{2}(-15-\sqrt{201})$

a)Generally the concave upward interval X is mathematically given as

$X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)$

$f''(x)$

b)Generally the concave downward interval Y is mathematically given as

$Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0 )$

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kykrilka [37]

Since the sequence is geometric, there is some constant $r$ such that the sequence is recursively given by

$a_n=ra_{n-1}$

By this definition, you can recursively substitute into the right hand side the definition for $a_{n-1},a_{n-2},\ldots$ to find an explicit formula for the $n$ th term.

$a_n=ra_{n-1}=r^2a_{n-2}=r^3a_{n-3}=\cdots=r^{n-1}a_1=-625r^{n-1}$

You know the second term, which means you can find $r$ :

$a_2=-625r^{2-1}\implies125=-625r\implies r=-\dfrac15$

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Sidana [21]

Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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