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3 years ago

8. Four Men working together can finish a house repair job in 12 days.

Mathematics

1 answer:

shutvik [7]3 years ago

8 0

Step-by-step explanation:

if 4 men did the job in 12 days , then each of the 4 men worked for 12 days , which means they worked for 4*12=48 days , so the job takes 48 work days to complete. If all men work at the same rate, then these other 6 will complete the job in

$\frac{48}{6} = 8 \frac{days}{man}$

Therefore, it’ll take the group of 6 men 8 days to complete the job.

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Write the expression as single natural logarithm <br> 3 in 3 + 2 in x ? Help needed 10 points.

leva [86]

Answer:

ln 27x²

Step-by-step explanation:

Using the rules of logarithms

• log $x^{n}$ ⇔ nlogx

• logx + logy ⇒ logxy

Given

3 ln 3 + 2 ln x, then

= ln 3³ + lnx²

= ln 27 + lnx² = ln 27x²

4 0

3 years ago

wind turbines generate large amounts of electricity from wind energy. the blades on a wind turbine can revolve 22 times per minu

lesya [120]

Answer:

1320 times

Step-by-step explanation:

Time = 1 h × (60 min/1 h) = 60 min

Revolutions = 60 min × (22 rev/1 min) = 1320 rev

The blades revolve 1320 times per hour.

5 0

3 years ago

P\left(x\right)=x^{3}+6x^{2}+5x-12

damaskus [11]

Answer: $P\left(x\right)=x^{3}+6x^{2}+5x-12$

Step-by-step explanation:

That is what she means! I don't know what you're asking but this may help the people that want to answer this question. :)

7 0

3 years ago

How do I write 36/10 as a decimal

saw5 [17]

The number 36 is really 36.000000... with infinite zeroes
so to divide by tan move the decimal one place to the left
<span>so 36/ 10= 3.6 </span>

5 0

4 years ago

Read 2 more answers

Determine whether the given system has a unique solution, no solution, or infinitely many solutions.

victus00 [196]

Answer:

The system has no solution.

Step-by-step explanation:

To find the solution to this system of linear equations

$\left\begin{array}{ccccc}-3x_1&+x_2&-2x_3&=&8&\\x_1&+5x_2&-x_3&=&4&\\-x_1&+11x_2&-4x_3&=&1&\end{array}\right$

First, state the problem in matrix form, this means, extracting only the numbers, and putting them in a box.

$\left[ \begin{array}{ccc|c} -3 & 1 & -2 & 8 \\\\ 1 & 5 & -3 & 4 \\\\ -1 & 11 & -4 & 1 \end{array} \right]$

This is called an augmented matrix. The word “augmented” refers to the vertical line, which we draw to remind ourselves where the equals sign belong

Next, transform the augmented matrix to the reduced row echelon form with the help of Row Operations.

Row Operation 1: multiply the 1st row by -1/3

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 1 & 5 & -1 & 4 \\\\ -1 & 11 & -4 & 1 \end{array} \right]$

Row Operation 2: add -1 times the 1st row to the 2nd row

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & \frac{16}{3} & - \frac{5}{3} & \frac{20}{3} \\\\ -1 & 11 & -4 & 1 \end{array} \right]$

Row Operation 3: add 1 times the 1st row to the 3rd row

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & \frac{16}{3} & - \frac{5}{3} & \frac{20}{3} \\\\ 0 & \frac{32}{3} & - \frac{10}{3} & - \frac{5}{3} \end{array} \right]$

Row Operation 4: multiply the 2nd row by 3/16

$\left[ \begin{array}{ccc|c} 1 & - \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & -15 \end{array} \right]$

Row Operation 5: add -32/3 times the 2nd row to the 3rd row

$\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & -15 \end{array} \right]$

Row Operation 6: multiply the 3rd row by -1/15

$\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & \frac{5}{4} \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

Row Operation 7: add -5/4 times the 3rd row to the 2nd row

$\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3} & \frac{2}{3} & - \frac{8}{3} \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

Row Operation 8: add 8/3 times the 3rd row to the 1st row

$\left[ \begin{array}{ccc|c} 1 &- \frac{1}{3} & \frac{2}{3} & 0 \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

Row Operation 9: add 1/3 times the 2nd row to the 1st row

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{16} & 0 \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

The reduced row echelon form of the augmented matrix is

$\left[ \begin{array}{cccc} 1 & 0 & \frac{9}{16} & 0 \\\\ 0 & 1 & - \frac{5}{16} & 0 \\\\ 0 & 0 & 0 & 1 \end{array} \right]$

which corresponds to the system

$\left\begin{array}{ccccc}x_1&&+\frac{9}{16} x_3&=&0&\\&1x_2&-\frac{5}{16}x_3&=&0&\\&&0&=&1&\end{array}\right$

Equation 3 cannot be solved, therefore, the system has no solution.

7 0

4 years ago