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3 years ago

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The specific gravity of a substance is the ratio of its weight to the weight of an equal volume of water. Kerosene weighs 0.82 g

ram per cubic centimeter and water weighs 1 gram per cubic centimeter. What is the specific gravity of kerosene?

2 answers:

netineya [11]3 years ago

8 0

Answer:

The specific gravity of kerosene is 0.82

Step-by-step explanation:

Let

x -----> the weight in grams of a cubic centimeter of Kerosene

y ----> the weight in grams of a cubic centimeter of water

z ---> the specific gravity of a substance

so

$z=\frac{x}{y}$

we have

$x=0.82\ g\\y=1\ g$

substitute the values

$z=\frac{0.82}{1}=0.82$

olga55 [171]3 years ago

7 0

Answer:

0.82

Step-by-step explanation:

The specific gravity of a substance is the ratio of its weight to the weight of an equal volume of water. Kerosene weighs 0.82 gram per cubic centimeter and water weighs 1 gram per cubic centimeter. Therefore, the specific gravity of kerosene is 0.82.

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Show all work to identify the asymptotes and end behavior of the function f(x)= 5x/x-25.

jeyben [28]

Considering the given function and the asymptote concept, we have that:

The vertical asymptote is of x = 25.

The horizontal asymptote is of y = 5.

The end behavior is that as $x \rightarrow \infty, f(x) \rightarrow 5$ .

<h3>What are the asymptotes of a function f(x)?</h3>

The vertical asymptotes are the values of x which are outside the domain, which in a fraction are the zeroes of the denominator.

The horizontal asymptote is the value of f(x) as x goes to infinity, as long as this value is different of infinity. Hence it also gives the end behavior of the function.

In this problem, the function is:

$f(x) = \frac{5x}{x - 25}$ .

The vertical asymptote is found as follows:

x - 25 = 0 -> x = 25.

The horizontal asymptote is found as follows:

$y = \lim_{x \rightarrow \infty} f(x) = \lim_{x \rightarrow \infty} \frac{5x}{x - 25} = \lim_{x \rightarrow \infty} \frac{5x}{x} = \lim_{x \rightarrow \infty} 5 = 5$ .

Hence the end behavior is that as $x \rightarrow \infty, f(x) \rightarrow 5$ .

More can be learned about asymptotes at brainly.com/question/16948935

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1 year ago

Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for each function.

Lunna [17]

Answer:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Step-by-step explanation:

GIven that:

$f(x) = 5e^{-x^2} cos (4x)$

The Maclaurin series of cos x can be expressed as :

$\mathtt{cos \ x = \sum \limits ^{\infty}_{n =0} (-1)^n \dfrac{x^{2n}}{2!} = 1 - \dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+... \ \ \ (1)}$

$\mathtt{e^{-2^x} = \sum \limits^{\infty}_{n=0} \ \dfrac{(-x^2)^n}{n!} = \sum \limits ^{\infty}_{n=0} (-1)^n \ \dfrac{x^{2n} }{x!} = 1 -x^2+ \dfrac{x^4}{2!} -\dfrac{x^6}{3!}+... \ \ \ (2)}$

From equation(1), substituting x with (4x), Then:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}- \dfrac{(4x)^6}{6!}+...}$

The first three terms of cos (4x) is:

$\mathtt{cos (4x) = 1 - \dfrac{(4x)^2}{2!}+ \dfrac{(4x)^4}{4!}-...}$

$\mathtt{cos (4x) = 1 - \dfrac{16x^2}{2}+ \dfrac{256x^4}{24}-...}$

$\mathtt{cos (4x) = 1 - 8x^2+ \dfrac{32x^4}{3}-... \ \ \ (3)}$

Multiplying equation (2) with (3); we have :

$\mathtt{ e^{-x^2} cos (4x) = ( 1- x^2 + \dfrac{x^4}{2!} ) \times ( 1 - 8x^2 + \dfrac{32 \ x^4}{3} ) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1+ (-8-1)x^2 + (\dfrac{32}{3} + \dfrac{1}{2}+8)x^4 + ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + (\dfrac{64+3+48}{6})x^4+ ...) }$

$\mathtt{ e^{-x^2} cos (4x) = ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

Finally , multiplying 5 with $\mathtt{ e^{-x^2} cos (4x) }$ ; we have:

The first three nonzero terms in the Maclaurin series is

$\mathbf{ 5e^{-x^2} cos (4x) }= \mathbf{ 5 ( 1 -9x^2 + \dfrac{115}{6}x^4+ ...) }$

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lesya [120]

Answer:

$y = 5x$ -- You

$y = 3 + 3x$ --- Your friend

Step-by-step explanation:

Given

You:

Trailhead implies the beginning so:

$Start = 0\ mile$

$Rate = 5miles/hour$

Your Friend

$Start = 3\ mile$

$Rate = 3miles/hour$

Required

Represent as an equation

From the question, we understand that:

$y = distance$

$x = number\ of\ hours$

So, the relationship between $distance\ and\ time\ is:$

$Distance = Start + Rate * Number\ of\ hours$

In terms $of\ x\ and\ y$ , we have:

$y = Start + Rate * x$

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For your friend, the equation becomes:

$y = 3 + 3 * x$

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2 years ago