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AveGali [126]
3 years ago
13

a daily newspaper has 10225 subscribers when it began publication. six years old later it has 8200 subscribers what is the avera

ge yearly rate of change in the number of subscribers for the six year period
Mathematics
1 answer:
seraphim [82]3 years ago
4 0

Answer:

The rate of change in number of subscriber for the six years is 3.62%

Step-by-step explanation:

Given as :

The initial subscriber of newspaper = p = 10225

The subscriber of newspaper after 6 years of publish = P = 8200

The time period = t = 6 years

Let The average yearly rate of change = r%

<u>Now, According to question</u>

The subscriber of newspaper after n years = initial subscriber × (1-\dfrac{\textrm rate}{100})^{\textrm time}

Or, P = p × (1-\dfrac{\textrm r}{100})^{\textrm t}

Or, 8200 = 10225 × (1-\dfrac{\textrm r}{100})^{\textrm 6}

Or, \dfrac{8200}{10225} =  (1-\dfrac{\textrm r}{100})^{\textrm 6}

Or, 0.80195 =  (1-\dfrac{\textrm r}{100})^{\textrm 6}

<u>Taking power \dfrac{1}{6} both side</u>

So, (0.80195 )^{\frac{1}{6}} = ((1 - \frac{r}{100} )^{6})^{\frac{1}{6}}

Or, 0.9638 = 1 - \dfrac{r}{100}

Or,  \dfrac{r}{100} = 1 - 0.9638

Or,  \dfrac{r}{100} = 0.0362

Or, r = 0.0362 × 100

i.e r = 3.62

So, The rate of change in subscriber = 3.62%

Hence, The rate of change in number of subscriber for the six years is 3.62% . Answer

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zhenek [66]

Step-by-step explanation:

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2 objects and their shadows create 2 similar right-angled triangles.

that means the angles are the same, and the side lengths of one triangle all correlate with the corresponding side lengths of the other triangle via the same factor.

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the length of Joel's shadow is then

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<h3>Washer method</h3>

Because the given region (R_{3}) has a look like a washer, we will apply the washer method to find the volume generated by rotating the given region about the specific line.

solution

We first find the value of x and y

y=2(x)^{\frac{1}{4} }

x=(\frac{y}{2} )^{4}

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v=\pi [\frac{1}{4} \frac{y^{3} }{3}  \int\limits^2_0 - \frac{1}{2^{8} }  \frac{y^{g} }{g} \int\limits^2_o\\v= \pi [\frac{1}{12} (2^{3} -0)-\frac{1}{2^{8}*9 } (2^{g} -0)]\\v= \pi [\frac{2}{3} -\frac{2}{g} ]\\v= \frac{4}{g} \pi

A similar question about finding the volume generated by a given region is answered here: brainly.com/question/3455095

6 0
1 year ago
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Morgarella [4.7K]

Answer:

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Step-by-step explanation:

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