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olchik [2.2K]
3 years ago
9

The wholesale price for a bookcase is $105. A certain furniture store marks up the wholesale price by 28% Find the price of the

bookcase in the furniture store.
Round your answer to the nearest cent, as necessary.
Mathematics
2 answers:
xxMikexx [17]3 years ago
8 0

Step-by-step explanation:

105 + 28%

find 28% of 105

of is multiply

28% × 105

28% as a decimal

0.28

0.28 × 105 = ??

then put answer into equation

105 + 28% ( ??)  =

answer in comments i will mark

Alex777 [14]3 years ago
7 0

Answer:

$134.40

Step-by-step explanation:

The markup is 28% of 105, or .28(105) = $29.40

Thus the retail price is $105 + $29.40 = $134.40

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10/16 of a pound is = 5/8 of a pound
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Jeremy bought a new truck for $32,000. The value of the truck after t years can be represented by the formula V = 32,000(.8)^t.
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6700=32000(0.8)^t
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Read 2 more answers
Find the reduced row echelon form of the following matrices and then give the solution to the system that is represented by the
GaryK [48]

Answer:

a)

Reduced Row Echelon:

\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&0&1&-4\end{array}\right]

Solution to the system:

x_3=-4\\x_2=-\frac{7}{4}x_3=7\\x_1=-\frac{1}{2}x_2=-\frac{7}{2}

b)

Reduced Row Echelon:

\left[\begin{array}{cccc}4&3&0&7\\0&0&2&-17\\0&0&2&-17\end{array}\right]

Solution to the system:  

x_3=-\frac{17}{2}\\x_1=\frac{7-3x_2}{4}

x_2 is a free variable, meaning that it has infinite possibilities and therefore the system has infinite number of solutions.

Step-by-step explanation:

To find the reduced row echelon form of the matrices, let's use the Gaussian-Jordan elimination process, which consists of taking the matrix and performing a series of row operations. For notation, R_i will be the transformed column, and r_i the unchanged one.

a) \left[\begin{array}{cccc}0&4&7&0\\2&1&0&0\\0&3&1&-4\end{array}\right]

Step by step operations:

1. Reorder the rows, interchange Row 1 with Row 2, then apply the next operations on the new rows:

R_1=\frac{1}{2}r_1\\R_2=\frac{1}{4}r_2

Resulting matrix:

\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&3&1&-4\end{array}\right]

2. Set the first row to 1

R_3=-3r_2+r_3

Resulting matrix:

\left[\begin{array}{cccc}1&1/2&0&0\\0&1&7/4&0\\0&0&1&-4\end{array}\right]

3. Write the system of equations:

x_1+\frac{1}{2}x_2=0\\x_2+\frac{7}{4}x_3=0\\x_3=-4

Now you have the  reduced row echelon matrix and can solve the equations, bottom to top, x_1 is column 1, x_2 column 2 and x_3 column 3:

x_3=-4\\x_2=-\frac{7}{4}x_3=7\\x_1=-\frac{1}{2}x_2=-\frac{7}{2}

b)

\left[\begin{array}{cccc}4&3&0&7\\8&6&2&-3\\4&3&2&-10\end{array}\right]

1. R_2=-2r_1+r_2\\R_3=-r_1+r_3

Resulting matrix:

\left[\begin{array}{cccc}4&3&0&7\\0&0&2&-17\\0&0&2&-17\end{array}\right]

2. Write the system of equations:

4x_1+3x_2=7\\2x_3=-17

Now you have the reduced row echelon matrix and can solve the equations, bottom to top, x_1 is column 1, x_2 column 2 and x_3 column 3:

x_3=-\frac{17}{2}\\x_1=\frac{7-3x_2}{4}

x_2 is a free variable, meaning that it has infinite possibilities and therefore the system has infinite number of solutions.

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