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sergejj [24]
3 years ago
13

Wanda opened a savings account 20 years ago with a deposit of $3,150.37. The account has an interest rate of 4.1% compounded dai

ly. How much interest has Wanda earned?
Mathematics
1 answer:
Aleksandr [31]3 years ago
3 0
\bf ~~~~~~ \stackrel{daily}{\textit{Continuously Compounding Interest Earned Amount}}\\\\
A=Pe^{rt}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
P=\textit{original amount deposited}\to& \$3150.37\\
r=rate\to 4.1\%\to \frac{4.1}{100}\to &0.041\\
t=years\to &20
\end{cases}
\\\\\\
A=3150.37e^{0.041\cdot 20}\implies A=3150.37e^{0.82}\implies A\approx 7152.91457

how much interest is there?  well, A - 3150.37.
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16. Find the final balance in an account with $730 invested at 3% annual simple interest for 4 years. (1 point) $735.48 $87.60 $
podryga [215]

Answer:

$817.60

Step-by-step explanation:

Simple interest has the formula:

Simple Interest = Principal * rate * time

here,

Principal amount is $730

The rate of interest is 3%, which is 3/100 = 0.03

The time length in years is given as 4

We find the interest amount first:

SI = P * r * t

SI = 730 * 0.03 * 4

SI = 87.6

We are asked to find FINAL BALANCE, which is PRINCIPAL + INTEREST

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7 0
3 years ago
X^2+4x+y^2-10y+20=30 find the center of the circle by completing the square
swat32

Answer:

a). Center of the circle = (-2, 5)

b). Equation of the line ⇒ y = -\frac{4}{5}x+\frac{58}{5}

Step-by-step explanation:

Equation of the circle is,

x² + 4x + y²- 10y + 20 = 30

a). [x² + 2(2)x + 4 - 4] + [y²- 2(5)y + 25] - 25 + 20 = 30

   [x² + 2(2)x + 4] - 4 + [y² - 2(5)y + 25] - 25 + 20 = 30

   (x + 2)² + (y - 5)²- 29 + 20 = 30

   (x + 2)² + (y - 5)²- 9 = 30

   (x + 2)² + (y - 5)² = 39

By comparing this equation with the standard equation of a circle,

    Center of the circle is (-2, 5).

b). A point (2, 10) lies on this circle.

    Slope of the line joining this point to the center (-2, 5),

    m_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

          = \frac{10-5}{2+2}

          = \frac{5}{4}

    Let the slope of the tangent which is perpendicular to this line is 'm_{2}'

    Then by the property of perpendicular lines,

          m_{1}\times m_{2}=-1

          \frac{5}{4}\times m_{2}=-1

                 m_{2}=-\frac{4}{5}

   Now the equation of the line passing though (2, 10) having slope m_{2}=-\frac{4}{5}

           y - y' = m_{2}(x-x')

           y - 10 = -\frac{4}{5}(x-2)

           y - 10 = -\frac{4}{5}x+\frac{8}{5}

                  y = -\frac{4}{5}x+\frac{8}{5}+10

                  y = -\frac{4}{5}x+\frac{58}{5}

Therefore, equation of the line will be, y = -\frac{4}{5}x+\frac{58}{5}

7 0
3 years ago
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