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3 years ago

G use part 1 of the fundamental theorem of calculus to find the derivative of the function. sqrt(4+7t)

Mathematics

1 answer:

uysha [10]3 years ago

6 0

Let y = √4+7t
then u= 4+7t
y=√u = u^½

du/dt= 7
dy/du = ½U^-½

dy/dt = du/dt • dy/du
= 7×½U^-½
= 7/2√U
= 7 / (2{√4+7t})

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Ulleksa [173]

This question uses trig

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3 years ago

Use the substitution method to solve the system of equations 5x-2y=9 3x+4y=-5

krek1111 [17]

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2 years ago

Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.

Aleks [24]

Answer:

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

Step-by-step explanation:

The actual Series is::

$\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$

The method we are going to use is comparison method:

According to comparison method, we have:

$\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n$

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

$=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}$

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}$

Now:

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}$

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

$\sum_{n=1}^{inf}\frac{1}{n^p}$

if p>1 then series converges. In oue case we have:

$\sum_{n=1}^{inf}b_n=\frac{1}{n^4}$

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

5 0

3 years ago

Calculate the frequency in hertz of photons of light with energy of 8.10 × 10-19 J.

stira [4]

The Planck–Einstein relation $E=hv$ gives that E, the photon energy, is equal to h, Planck's constant, times <span>ν</span>, the frequency.

We can solve the equation in terms of frequency to get
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We substitute our given energy, $8.10 \times 10^{-19} J$ , and Planck's constant, $6.626 \times 10^{-34} Js$ .

$v= \dfrac{8.10 \times 10^{-19} J}{6.626 \times 10^{-34} Js}=1.22 \times 10^{15} s^{-1}=1.22 \times 10^{15} Hz$

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3 years ago

9.64 rounded to the nearest tenth

Ivanshal [37]

Hello,

9.64 rounded to the nearest tenth is 9.6.

Have A Good Day.

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3 years ago

Read 2 more answers