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taurus [48]
3 years ago
15

six cells phones weigh 46.2 ounces. fourteen cell phones weigh 107.8 ounces. does this represent a proportional relationship? ex

plain. if so, state the constant of proportionality.
Mathematics
2 answers:
stealth61 [152]3 years ago
7 0
Well first we must see if they are proportional.

46.2/6 = 10.7.8/14
7.7 = 7.7

They are proportional. The rate of proportion is 7.7 ounces per cellphone. Hope this helps!
sattari [20]3 years ago
7 0

Answer:

<h2>They have a proportional relationship, with a constant of proportionality of 7.7</h2>

Step-by-step explanation:

<em>A proportional relationship refers to the equivalence between two reasons, or two fractions.</em>

So, in this case, we have the two reasons.

<h3>First reason: </h3>

\frac{46.2 \ ounces }{6 \ cellphones}

<h3>Second reason:</h3>

\frac{107.8 \ ounces}{14 \ cellphones}

Now, to find if these reasons are proportional we just have to equalize them and find if both fractions have the same result:

\frac{46.2 \ ounces }{6 \ cellphones}=\frac{107.8 \ ounces}{14 \ cellphones}\\7.7=7.7

<em>As you can see, both fractions have the same result, this means that those reasons have a proportional relationship, and the constant of proportionality is 7.7.</em>

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(a) Use the power series expansions for ex, sin x, cos x, and geometric series to find the first three nonzero terms in the powe
Fofino [41]

Answer:

a) \mathbf{4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!}  ...}

b)  See Below for proper explanation

Step-by-step explanation:

a) The objective here  is to Use the power series expansions for ex, sin x, cos x, and geometric series to find the first three nonzero terms in the power series expansion of the given function.

The function is e^x + 3 \ cos \ x

The expansion is of  e^x is e^x = 1 + \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + ...

The expansion of cos x is cos \ x = 1 - \dfrac{x^2}{2!}+ \dfrac{x^4}{4!}- \dfrac{x^6}{6!}+ ...

Therefore; e^x + 3 \ cos \ x  = 1 + \dfrac{x}{1!}+ \dfrac{x^2}{2!}+ \dfrac{x^3}{3!} + ... 3[1 - \dfrac{x^2}{2!}+ \dfrac{x^4}{4!}- \dfrac{x^6}{6!}+ ...]

e^x + 3 \ cos \ x  = 4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!} + \dfrac{x^3}{3!}+ ...

Thus, the first three terms of the above series are:

\mathbf{4 + \dfrac{x}{1!}- \dfrac{2x^2}{2!}  ...}

b)

The series for e^x + 3 \ cos \ x is \sum \limits^{\infty}_{x=0} \dfrac{x^x}{n!} +  3 \sum \limits^{\infty}_{x=0} ( -1 )^x  \dfrac{x^{2x}}{(2n)!}

let consider the series; \sum \limits^{\infty}_{x=0} \dfrac{x^x}{n!}

|\frac{a_x+1}{a_x}| = | \frac{x^{n+1}}{(n+1)!} * \frac{n!}{x^x}| = |\frac{x}{(n+1)}| \to 0 \ as \ n \to \infty

Thus it converges for all value of x

Let also consider the series \sum \limits^{\infty}_{x=0}(-1)^x\dfrac{x^{2n}}{(2n)!}

It also converges for all values of x

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