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STatiana [176]
3 years ago
9

$20.600 at 8% for 2 years

Mathematics
1 answer:
balu736 [363]3 years ago
6 0

Answer:

$3.296

Step-by-step explanation:

Simple interest=PRT/100

P=Principal=$20.600

R=rate=8%

T=time=2yrs

=20.600×8/100×2

=$3.296

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Perform long division on the​ integrand, write the proper fraction as a sum of partial​ fractions, and then evaluate the integra
worty [1.4K]

Answer:

x^2+4x -\frac{1}{2}  lnx  + \frac{2}{13}  ln(x-2) + C

Step-by-step explanation:

Given the integrand \int\limits{\dfrac{2x^3 - 2x + 1}{x^2-2x} } \, dx, before evaluating the integral function, we will need to simplify the function first by applying long division as shown in the attachment.

Hence the partial form of the function \dfrac{2x^3 - 2x + 1}{x^2-2x} } = 2x+4 + \frac{6x+1}{x^2-2x}

Integrating its partial sum

\int\limits \dfrac{2x^3 - 2x + 1}{x^2-2x} }dx  = \int\limits  (2x+4 + \frac{6x+1}{x^2-2x})\ dx\\\\= \int\limits {2x} \, dx + \int\limits {4} \, dx + \int\limits {\frac{6x+1}{x^2-2x} \, dx\\ = \frac{2x^2}{2}+4x -\frac{1}{2}  \int\limits{\frac{1}{x} } \, dx  + \frac{2}{13}  \int\limits{\frac{1}{x-2} } \, dx

=  \frac{2x^2}{2}+4x -\frac{1}{2}  lnx  + \frac{2}{13}  ln(x-2) + C

= x^2+4x -\frac{1}{2}  lnx  + \frac{2}{13}  ln(x-2) + C

<em>NB: Find the partial sum calculation also in the attachment. </em>

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3 years ago
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NEED HELP NOW !!!!!!! The ratio of teachers to students on a field trip to the art museum is supposed to be 1 : 5. Which group d
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Step-by-step explanation:

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3 years ago
What is an equation of the line that passes through the point (-2,7) and is perpendicular to the line x-6y=42 ?
Natali5045456 [20]

Answer:

The equation of the line that passes through the point (-2,7) and is perpendicular to the line x-6y=42 is

6x+y=-5

Step-by-step explanation:

Given:  

Let,  

point A( x₁ , y₁) ≡ ( -2 , 7)

To Find:  

Equation of Line  that passes through the point (-2,7) and is perpendicular to the line x-6y=42=?  

Solution:  

x-6y=42    ..................Given

which can be written as

y=mx+c

Where m is the slope of the line

∴ y=\dfrac{x}{6}-7

On Comparing we get

Slope = m = \dfrac{1}{6}

The Required line is Perpendicular to the above line.

So,

Product of slopes = - 1

m\times m_{1}=-1\\Substituting\ m\\ \dfrac{1}{6} m_{1}=-1\\\\m_{1}=-6

Slope of the required line is -6

Equation of a line passing through a points A( x₁ , y₁) and having slope m is given by the formula,  

i.e equation in point - slope form

(y-y_{1})=m(x-x_{1})

Now on substituting the slope and point A( x₁ , y₁) ≡ ( -2, 7) and slope = -6 we get

(y-7)=-6(x--2)=-6(x+2)=-6x-12\\\\\therefore 6x+y=-5.......is\ the required\ equation\ of\ the\ line

The equation of the line that passes through the point (-2,7) and is perpendicular to the line x-6y=42 is

6x+y=-5

3 0
3 years ago
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