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olga55 [171]
3 years ago
12

What is 6.12 as a fraction

Mathematics
2 answers:
timurjin [86]3 years ago
6 0
12 6 100 612/100 I hope this helps
liubo4ka [24]3 years ago
3 0
The six is the whole number, and twelve is the numerator. The reason why it's a numerator is because it's not whole. The reason why the denominator would be 100 is because the 2 is in the hundredth's place.
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Has a slope of -2/5 and goes through (-10,-1)​
Aleonysh [2.5K]

Answer:

Below in bold.

Step-by-step explanation:

Using the point-slope form of a straight line equation:

y - y1 = m(x - x1)

y - (-1)) = -2/5(x - -10)

y + 1 = -2/5(x + 10)

y  + 1 = -2/5x - 4

y = -2/5x - 5.

In standard form this is:

2x + 5y = -25.

7 0
2 years ago
Write 2.18 as a mixed number in simplest form
bezimeni [28]
It would be: 2 18/100 = 2 9/50

So, your final answer is 2 9/50

Hope this helps!
8 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
Find the equation of a line that is parallel to 2x-8y=16 and passes through the point (0,3).
denis23 [38]
-8y = -2x + 16
y = 1/4x - 2
y = 1/4x + 3
3 0
3 years ago
The local parts shop buys a machine that costs $500,000. It's value depreciates exponentially each year by 10%.
Musya8 [376]
<span>After 5 years, it's value will be about $294,300</span>
3 0
3 years ago
Read 2 more answers
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