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VashaNatasha [74]
3 years ago
9

Kate borrowed $400 from her friend. She will repay this amount plus 3 percent simple interest after one year. What is the total

amount she will pay her friend?
Mathematics
1 answer:
tekilochka [14]3 years ago
6 0
The answer is $412.

Let's first calculate simple interest. Simple interest (I) can be expressed as:
I = P * r * t
P - principal
r - rate
t - time period

It is given:
I = ?
P = $400
r = 3% = 0.03
t = 1 year

Therefore:
I = P * r * t = 400 * 0.03 * 1 = 12

The total amount Kate will repay is the principal amount (P) plus 3% simple interest (I):
P + I = 400 + 12 = $412
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Find the area of the regular decagon if the apothem is 2.8ft and a side is 1.8ft
geniusboy [140]
DECA=10, decagon = 10 sides.

since each side is 1.8 long, all 10 sides will be 10*1.8 or 18 units long, so that'd be the perimeter then,

\bf \textit{area of a regular polygon}\\\\
A=\cfrac{1}{2}ap~~
\begin{cases}
a=apothem\\
p=perimeter\\
------\\
a=2.8\\
p=18
\end{cases}\implies A=\cfrac{1}{2}(2.8)(18)
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3 years ago
Find a solution of x dy dx = y2 − y that passes through the indicated points. (a) (0, 1) y = (b) (0, 0) y = (c) 1 6 , 1 6 y = (d
Leni [432]
Answers: 

(a) y = \frac{1}{1 - Cx}, for any constant C

(b) Solution does not exist

(c) y = \frac{256}{256 - 15x}

(d) y = \frac{64}{64 - 15x}

Explanations:

(a) To solve the differential equation in the problem, we need to manipulate the equation such that the expression that involves y is on the left side of the equation and the expression that involves x is on the right side equation.

Note that

 x\frac{dy}{dx} = y^2 - y
\\
\\ \indent xdy = \left ( y^2 - y \right )dx
\\
\\ \indent \frac{dy}{y^2 - y} = \frac{dx}{x}
\\
\\ \indent \int {\frac{dy}{y^2 - y}} = \int {\frac{dx}{x}} 
\\
\\ \indent \boxed{\int {\frac{dy}{y^2 - y}} = \ln x + C_1}      (1)

Now, we need to evaluate the indefinite integral on the left side of equation (1). Note that the denominator y² - y = y(y - 1). So, the denominator can be written as product of two polynomials. In this case, we can solve the indefinite integral using partial fractions.

Using partial fractions:

\frac{1}{y^2 - y} = \frac{1}{y(y - 1)} = \frac{A}{y - 1} + \frac{B}{y}
\\
\\ \indent \Rightarrow \frac{1}{y^2 - y} = \frac{Ay + B(y-1)}{y(y - 1)} 
\\
\\ \indent \Rightarrow \boxed{\frac{1}{y^2 - y} = \frac{(A+B)y - B}{y^2 - y} }      (2)

Since equation (2) has the same denominator, the numerator has to be equal. So,

1 = (A+B)y - B
\\
\\ \indent \Rightarrow (A+B)y - B = 0y + 1
\\
\\ \indent \Rightarrow \begin{cases}
 A + B = 0
& \text{(3)}\\-B = 1
 & \text{(4)}   \end{cases}

Based on equation (4), B = -1. By replacing this value to equation (3), we have

A + B = 0
A + (-1) = 0
A + (-1) + 1 = 0 + 1
A = 1 

Hence, 

\frac{1}{y^2 - y} = \frac{1}{y - 1} - \frac{1}{y}

So,

\int {\frac{dy}{y^2 - y}} = \int {\frac{dy}{y - 1}} - \int {\frac{dy}{y}} 
\\
\\ \indent \indent \indent \indent = \ln (y-1) - \ln y
\\
\\ \indent  \boxed{\int {\frac{dy}{y^2 - y}} = \ln \left ( \frac{y-1}{y} \right ) + C_2}

Now, equation (1) becomes

\ln \left ( \frac{y-1}{y} \right ) + C_2 = \ln x + C_1
\\
\\ \indent \ln \left ( \frac{y-1}{y} \right ) = \ln x + C_1 - C_2
\\
\\ \indent  \frac{y-1}{y} = e^{C_1 - C_2}x
\\
\\ \indent  \frac{y-1}{y} = Cx, \text{ where } C = e^{C_1 - C_2}
\\
\\ \indent  1 - \frac{1}{y} = Cx
\\
\\ \indent \frac{1}{y} = 1 - Cx
\\
\\ \indent \boxed{y = \frac{1}{1 - Cx}}
       (5)

At point (0, 1), x = 0, y = 1. Replacing these values in (5), we have

y = \frac{1}{1 - Cx}
\\
\\ \indent 1 = \frac{1}{1 - C(0)} = \frac{1}{1 - 0} = 1



Hence, for any constant C, the following solution will pass thru (0, 1):

\boxed{y = \frac{1}{1 - Cx}}

(b) Using equation (5) in problem (a),

y = \frac{1}{1 - Cx}   (6)

for any constant C.

Note that equation (6) is called the general solution. So, we just replace values of x and y in the equation and solve for constant C.

At point (0,0), x = 0, y =0. Then, we replace these values in equation (6) so that 

y = \frac{1}{1 - Cx}
\\
\\ \indent 0 = \frac{1}{1 - C(0)} = \frac{1}{1 - 0} = 1

Note that 0 = 1 is false. Hence, for any constant C, the solution that passes thru (0,0) does not exist.

(c) We use equation (6) in problem (b) and because equation (6) is the general solution, we just need to plug in the value of x and y to the equation and solve for constant C. 

At point (16, 16), x = 16, y = 16 and by replacing these values to the general solution, we have

y = \frac{1}{1 - Cx}
\\
\\ \indent 16 = \frac{1}{1 - C(16)} 
\\ 
\\ \indent 16 = \frac{1}{1 - 16C}
\\
\\ \indent 16(1 - 16C) = 1
\\ \indent 16 - 256C = 1
\\ \indent - 256C = -15
\\ \indent \boxed{C = \frac{15}{256}}




By replacing this value of C, the general solution becomes

y = \frac{1}{1 - Cx}
\\
\\ \indent y = \frac{1}{1 - \frac{15}{256}x} 
\\ 
\\ \indent y = \frac{1}{\frac{256 - 15x}{256}}
\\
\\
\\ \indent \boxed{y = \frac{256}{256 - 15x}}





This solution passes thru (16,16).

(d) We do the following steps that we did in problem (c):
        - Substitute the values of x and y to the general solution.
        - Solve for constant C

At point (4, 16), x = 4, y = 16. First, we replace x and y using these values so that 

y = \frac{1}{1 - Cx} 
\\ 
\\ \indent 16 = \frac{1}{1 - C(4)} 
\\ 
\\ \indent 16 = \frac{1}{1 - 4C} 
\\ 
\\ \indent 16(1 - 4C) = 1 
\\ \indent 16 - 64C = 1 
\\ \indent - 64C = -15 
\\ \indent \boxed{C = \frac{15}{64}}

Now, we replace C using the derived value in the general solution. Then,

y = \frac{1}{1 - Cx} \\ \\ \indent y = \frac{1}{1 - \frac{15}{64}x} \\ \\ \indent y = \frac{1}{\frac{64 - 15x}{64}} \\ \\ \\ \indent \boxed{y = \frac{64}{64 - 15x}}
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Ugo [173]
There are 4 eggs per hen
What you have to do is
24 hens/ 96 eggs
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