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Ierofanga [76]
3 years ago
15

zoe wrote 3/5 of her essay in 30 minutes. assuming she wrote the paper at a constant rate the entire time, which expression can

be used to determine the fraction of her essay she wrote each minute?
Mathematics
2 answers:
vodomira [7]3 years ago
6 0
3/5 in 30 mins, 5*30=150, then she would write 3/150 or 1/50 each minute
alexandr1967 [171]3 years ago
3 0

Answer: Ok, so we know that zoe wrote 3/5 of her essay in 30 minutes.

Now, assuming she wrote the paper at a constant rate, then if we divide those 3/5 in 3, we get 1/5 + 1/5 + 1/5, t we also divide the time by the same number, then:

1/5 + 1/5 + 1/5 = 10m +10m +10m

so she wrote a 1/5 of her essay in 10 minutes.

if we divide by 10 this time, we get 1/50 = 1 minute.

then she wrote 1/50 of her essay per minute.

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3 years ago
Solve the system of equations:x + 3y - z = -4 2x - y + 2z = 13 3x - 2y - z = -9
tatiyna

Answer:

The solution to the system of equations is

\begin{gathered} x=\frac{179}{13} \\  \\ y=-\frac{279}{39} \\  \\ z=-\frac{48}{13} \end{gathered}

Explanation:

Giving the system of equations:

\begin{gathered} x+3y-z=-4\ldots\ldots\ldots\ldots\ldots\ldots..........\ldots\ldots\ldots\ldots.\ldots\text{.}\mathrm{}(1) \\ 2x-y+2z=13\ldots\ldots...\ldots\ldots\ldots\ldots..\ldots..\ldots\ldots\ldots\ldots\ldots.(2) \\ 3x-2y-z=-9\ldots\ldots\ldots.\ldots\ldots\ldots\ldots....\ldots\ldots.\ldots\ldots\ldots\text{.}\mathrm{}(3) \end{gathered}

To solve this, we need to first of all eliminate one variable from any two of the equations.

Subtracting (2) from twice of (1), we have:

5y-4z=-21\ldots\ldots\ldots\ldots\ldots.\ldots.\ldots..\ldots..\ldots\ldots.\ldots..\ldots\text{...}\mathrm{}(4)

Subtracting (3) from 3 times (1), we have

3y-5z=-3\ldots\ldots...\ldots\ldots..\ldots\ldots\ldots\ldots\ldots.\ldots\ldots\ldots\ldots\ldots..\ldots\ldots(5)

From (4) and (5), we can solve for y and z.

Subtract 5 times (5) from 3 times (4)

\begin{gathered} 13z=-48 \\  \\ z=-\frac{48}{13} \end{gathered}

Using the value of z obtained in (5), we have

\begin{gathered} 3y-5(-\frac{48}{13})=-3 \\  \\ 3y+\frac{240}{13}=-3 \\  \\ 3y=-3-\frac{240}{13} \\  \\ 3y=-\frac{279}{13} \\  \\ y=-\frac{279}{39} \end{gathered}

Using the values obtained for y and z in (1), we have

\begin{gathered} x+3(-\frac{279}{39})-(-\frac{48}{13})=-4 \\  \\ x-\frac{279}{13}+\frac{48}{13}=-4 \\  \\ x-\frac{231}{13}=-4 \\  \\ x=-4+\frac{231}{13} \\  \\ x=\frac{179}{13} \end{gathered}

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