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3 years ago

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The average human heart beats 1.15 \cdot 10^51.15⋅10 5 1, point, 15, dot, 10, start superscript, 5, end superscript times per da

y. There are 3.65 \cdot 10^23.65⋅10 2 3, point, 65, dot, 10, start superscript, 2, end superscript days in one year. How many times does the heart beat in one year?

2 answers:

andriy [413]3 years ago

4 0

Answer:

$4.1975\cdot 10^{7}$

Step-by-step explanation:

We are told that the average human heart beats $1.15\cdot 10^{5}$ times per day and there are $3.65\cdot 10^{2}$ days in one year.

To find number of heart beats in one year we will multiply number of heart beats in one day by number of days in one year.

$1.15\cdot 10^{5}\times 3.65\cdot 10^{2}$

Now we will solve this problem using exponent properties.

$1.15 \times 3.65\cdot 10^{2+5}$

$1.15 \times 3.65\cdot 10^{7}$

$4.1975\cdot 10^{7}$

Our answer is in scientific notation we can represent it in standard form as $4.1975\cdot 10^{7}=4.1975*10000000=41975000$ times.

Therefore, average human heart beats in one year $4.1975\cdot 10^{7}$ or 41975000 times.

Lesechka [4]3 years ago

3 0

Answer:

4.2*10^7

Step-by-step explanation:

Khan Academy

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SOLUTION We observe that f '(x) = -1 / (1 + x2) and find the required series by integrating the power series for -1 / (1 + x2).

Ann [662]

Answer:

Required series is:

$\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

Step-by-step explanation:

Given that

$f'(x) = -\frac{1}{1 + x^{2}}$ ---(1)

We know that:

$\frac{d}{dx}(tan^{-1}x)=\frac{1}{1+x^{2}}$ ---(2)

Comparing (1) and (2)

$f'(x)=-(tan^{-1}x)$ ---- (3)

Using power series expansion for $tan^{-1}x$

$f'(x)=-tan^{-1}x=-\int {\frac{1}{1+x^{2}} \, dx$

$= -\int{ \sum\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx$

$= -\sum{ \int\limits^{ \infty}_{n=0} (-1)^{n}x^{2n}} \, dx$

$=-[c+\sum\limits^{ \infty}_{n=0} (-1)^{n}\frac{x^{2n+1}}{2n+1}]$

$=C+\sum\limits^{ \infty}_{n=0} (-1)^{n+1}\frac{x^{2n+1}}{2n+1}$

$=C-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

as

$tan^{-1}(0)=0 \implies C=0$

Hence,

$\int{\frac{-1}{1+x^{2}} \, dx =-x+\frac{x^{3}}{3}-\frac{x^{5}}{5}+\frac{x^{7}}{7}+.....$

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Vassil rounded to the nearest half to estimate the product of Five-eighths and StartFraction 8 over 9 EndFraction. How do the es

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The correct answer is option B which is the actual product is about StartFraction 1 over 18 EndFraction less than the estimate.

<h3>What are fractions?</h3>

The fraction is defined as the division of the whole part into an equal number of parts. here we have the following information Vassil rounded to the nearest half to estimate the product of Five-eighths and StartFraction 8 over 9 EndFraction.

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Thus the product of the fraction estimated in the nearest half is,

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Therefore the correct answer is option B which is the actual product is about StartFraction 1 over 18 EndFraction less than the estimate.

To know more about fractions follow

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#SPJ1

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