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3 years ago

What fraction is equivalent to 16.6 repeating

Mathematics

2 answers:

irina [24]3 years ago

6 0

The answer is 16 2/3 ?

Jobisdone [24]3 years ago

4 0

Answer:

It is 16.666…

let 16.666 be x :

$x = 16.666 - - - (i) \\ 100x = 1666.666 - - - (ii)$

subtract (i) from (ii):

$99x = 1650 \\ \\ x = \frac{1650}{99} = { \boxed{ \boxed{ \frac{50}{3} }}}$

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Which equation has x = -2 as the solution?

Vera_Pavlovna [14]

The answer is 2x + 8 = 4

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3 years ago

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What will be the value in twenty years of $1000 invested at the end of each year for the next twenty years?

denis23 [38]

The formula for yearly compound interest is A(1 + r)^t

Where is A = 1000, r = 8%, t = 20

$1000(1 + .08)^{20} = 4660.96$

The answer is $4660.96

3 0

3 years 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

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

7 0

3 years ago

Using logarithm table evaluate the <img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B%20%5Csqrt%5B3%20%5D%7B1.376%7D%20%7D%7B%20

DedPeter [7]

Answer: 3.845312455 just put it in a scientific calculator

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3 years ago

Calculate 19.25tons=___lbs.

Ket [755]

Answer:

19.25 tons = 38500 lbs

Step-by-step explanation:

We are to convert the following given amount of tons in pounds.

We know that, 1 ton = 2000 pounds. So using the ration method, we can convert 19.25 tons into pounds.

$\frac{1 ton}{19.25 tons} =\frac{2000 lbs}{x}$

$x = 2 0 0 0 \times 1 9 . 2 5$

$x = 3 8 5 0 0 lbs$

Therefore, 19.25 tons = 38500 lbs.

7 0

3 years ago

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