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EastWind [94]
2 years ago
15

How do I get the result for 1/8 out of 40?

Mathematics
2 answers:
alekssr [168]2 years ago
7 0
Answer:5
Step by step
You divide your number by the denominator and the time it by the numerator
velikii [3]2 years ago
3 0

Answer:

5

Step-by-step explanation:

You divide your number by the denominator and then times it by the numerator

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Math question please help!! thank you if you help
algol [13]
These solutions can be called “roots”, “zeroes”, and “x-intercepts”.
6 0
2 years ago
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. XYZ has vertices X (–5, 2), Y (0, –4), and Z (3, 3). What are the vertices of the image of X’Y’Z’ under the translation (x, y)
Katen [24]

Answer:

X' = (2, -3)

Y' = (7, -9)

Z' = (10, -2)

Step-by-step explanation:

to find the translation of each point substitute the x and y with the the numbers

eg; X ( -5,2) the x is 5 and the y is 2

so -5 + 7 = 2 (the x value)

2 - 5 = -3 ( the y value)

6 0
3 years ago
Find three consecutive odd numbers whose sum is 303
CaHeK987 [17]
n;\ n+2;\ n+4-three\ consecutive\ odd\ numbers\\\\n+(n+2)+(n+4)=303\\\\3n+6=303\\\\3n=303-6\\\\3n=297\ \ \ \ /:3\\\\n=99\\\\Answer:99;\ 101;\ 103.
8 0
3 years ago
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A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund f
Andrews [41]

Answer:

A = \frac{P}{r}\left( e^{rt} -1 \right)

Step-by-step explanation:

This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

                                                \frac{dA}{rA+P} = dt

Integration on both sides gives

                                            \int \frac{dA}{rA+P} = \int  dt

where c is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

                               \int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c

Therefore,

                                        \frac{1}{r} \ln |rA+P| = t+c

Multiply both sides by r.

                               \ln |rA+P| = rt+c_1, \quad c_1 := rc

By taking exponents, we obtain

      e^{\ln |rA+P|} = e^{rt+c_1} \implies  |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}

Isolate A.

                 rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}

Since A = 0  when t=0, we obtain an initial condition A(0) = 0.

We can use it to find the numeric value of the constant c.

Substituting 0 for A and t in the equation gives

                         0 = \frac{C}{r}e^{0} - \frac{P}{r} \implies \frac{P}{r} = \frac{C}{r} \implies C=P

Therefore, the solution of the given differential equation is

                                   A = \frac{P}{r}e^{rt} - \frac{P}{r} = \frac{P}{r}\left( e^{rt} -1 \right)

4 0
3 years ago
Last Question On The Quiz Middle School Math Medium/Easy
VikaD [51]

Answer:

B

Step-by-step explanation:

6 0
3 years ago
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