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3 years ago

Suppose that 13 inches of wire costs 52 cents. At the same rate, how much (in cents) will 34 inches of wire cost.

Mathematics

1 answer:

dimulka [17.4K]3 years ago

8 0

Find price per inch:

0.52 / 13 = 0.04 per inch.

Multiply cost per inch by inches:

0.04 x 34 = 1.36

It will cost $1.36 for 34 inches.

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Matt is a software engineer writing a script involving 6 tasks. Each must be done one after the other. Let ti be the time for th

Masteriza [31]

Answer:

Let $t_i$ be the time for the $i$ th task.

We know these times have a certain structure:

Any 3 adjacent tasks will take half as long as the next two tasks.

In the form of an equations we have

$t_1+t_2+t_3=\frac{1}{2}t_4+\frac{1}{2}t_5 \\\\t_2+t_3+t_4=\frac{1}{2}t_5+\frac{1}{2}t_6$

The second task takes 1 second $t_2=1$

The fourth task takes 10 seconds $t_4=10$

So, we have the following system of equations:

$t_1+t_2+t_3-\frac{1}{2}t_4-\frac{1}{2}t_5=0 \\\\t_2+t_3+t_4-\frac{1}{2}t_5-\frac{1}{2}t_6=0\\\\t_2=1\\\\t_4=10$

a) An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.

Here is the augmented matrix for this system.

$\left[ \begin{array}{cccccc|c} 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 & 0 \\\\ 0 & 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]$

b) To reduce this augmented matrix to reduced echelon form, you must use these row operations.

Subtract row 2 from row 1 $\left(R_1=R_1-R_2\right)$ .
Subtract row 2 from row 3 $\left(R_3=R_3-R_2\right)$ .
Add row 3 to row 2 $\left(R_2=R_2+R_3\right)$ .
Multiply row 3 by −1 $\left({R}_{{3}}=-{1}\cdot{R}_{{3}}\right)$ .
Add row 4 multiplied by $\frac{3}{2}$ to row 1 $\left(R_1=R_1+\left(\frac{3}{2}\right)R_4\right)$ .
Subtract row 4 from row 3 $\left(R_3=R_3-R_4\right)$ .

Here is the reduced echelon form for the augmented matrix.

$\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right]$

c) The additional rows are

$\begin{array}{ccccccc} 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right$

and the augmented matrix is

$\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right]$

d) To solve the system you must use these row operations.

Subtract row 1 from row 6 $\left(R_6=R_6-R_1\right)$ .
Subtract row 2 from row 6 $\left(R_6=R_6-R_2\right)$ .
Subtract row 3 from row 6 $\left(R_6=R_6-R_3\right)$ .
Swap rows 5 and 6.
Add row 5 to row 3 $\left(R_3=R_3+R_5\right)$ .
Multiply row 5 by 2 $\left(R_5=\left(2\right)R_5\right)$ .
Subtract row 6 multiplied by 1/2 from row 1 $\left(R_1=R_1-\left(\frac{1}{2}\right)R_6\right)$ .
Add row 6 multiplied by 1/2 to row 3 $\left(R_3=R_3+\left(\frac{1}{2}\right)R_6\right)$ .

$\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & 0 & 0 & 44 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 1 & 0 & 90 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \end{array} \right]$

The solutions are: $(t_1,...,t_6)=(5,1,44,10,90,20)$ .

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3 years ago

David and Karen are building a treehouse in the shape of a rectangular prism for their daughter.If the treehouse is going to 5 f

Serga [27]

The space left inside the tree is 300 cubic feet.

David and Karen have to paint 275 square feet on the outside.

Explanation:

It is given that the length of the tree house is 7.5 feet

The width of the tree house is 8 feet

The height of the tree house is 5 feet

The tree house is in the shape of a rectangular prism.

The volume of the rectangular prism is given by

$\text {Volume}=\text {length } \times \text {width} \times \text {height}$

Substituting the values, we have,

$Volume$=7.5 \times 8 \times 5$\\Volume $=300$$

Thus, the volume of the rectangular prism is 300 cubic feet

Hence, the space left inside the tree is 300 cubic feet.

The area they have to paint on the outside can be determined using the formula for surface area of the prism .

$Area=2(w l+h l+h w)$

Substituting the values, we get,

$Area=2[(8*7.5)+(5*7.5)+(5*8)]$

Multiplying the terms within the bracket, we get,

$Area=2(60+37.5+40)$

Adding the terms, we have,

$Area=2 \times 137.5$

Multiplying, we get,

$Area =275$

Thus, David and Karen have to paint 275 square feet on the outside.

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3 years ago

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I am Lyosha [343]

$1800|2\\.\ 900|2\\.\ 450|2\\.\ 225|5\\.\ \ 45|5\\.\ \ \ 9|3\\.\ \ \ 3|3\\.\ \ \ 1|\\\\1880=2\cdot2\cdot2\cdot5\cdot5\cdot3\cdot3=2^3\cdot5^2\cdot3^2\\\\Answer:\ \boxed{2^3\cdot3^2\cdot5^2}$

$\sqrt[3]{320}\\\\320|2\\160|2\\.\ 80|2\\.\ 40|2\\.\ 20|2\\.\ 10|2\\.\ \ 5|5\\.\ \ 1|\\\\320=2\cdot2\cdot2\cdot2\cdot2\cdo2\cdot5=2^3\cdot2^3\cdot5=2^3\cdot2^3\cdot\\\\\sqrt[3]{320}=\sqrt[3]{2^3\cdot2^3\cdot5}=\sqrt[3]{2^3}\cdot\sqrt[3]{2^3}\cdot\sqrt[3]{5}=2\cdot2\sqrt[3]{5}=4\sqrt[3]5\\\\Answer:\ \boxed{\sqrt[3]{320}=4\sqrt[3]{5}}$

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LuckyWell [14K]

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