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Katyanochek1 [597]
3 years ago
12

Suppose that the cost of a paving stone is $2.50, plus $0.15 for every 4 cubic inches of concrete.

Mathematics
1 answer:
olganol [36]3 years ago
8 0

Answer:

6 different sized paving stones,$16

Complete question:

What if the 360 cubic-inch paving stones are 4 inches thick and any whole number length and width are possible? How many different paving stones could be made? Suppose that the cost of having stone is $2.50, plus $0.15 for every 4 cubic inches of concrete how much would each paving stone cost?

Step-by-step explanation:

V= B x h

B= V / h=> 360 / 4  

B= 90 sq inch

Considering the factors of 90: 1,2,3,5,6,9,10,15,18,30,45,90

Now, make table with base height and volume for each pair of factors. (see figure in the attachment)

We'll have 6 different sized paving stones.

As each stone has a vol of 360 inches³. Diving by 4 in order to find how many 4 inch³ per stone

Concrete=$0.15 x (360/4) => $0.15 x 90

Concrete= $13.5

The cost of the stone plus the concrete will be:

cost= $2.50 + concrete

cost= $2.50 + $13.5

cost=$16

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The products of AB and BA is given by

AB=\left[\begin{array}{cc}21&-6\\ 9&3\end{array}\right]

BA=\left[\begin{array}{cc}9&-1\\-18&15\end{array}\right]

Step-by-step explanation:

Given the matrices A=\left[\begin{array}{cc}6&-5\\-3&-4\end{array}\right] and

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To find the product AB and BA

AB=\left[\begin{array}{cc}6&-5\\-3&-4\end{array}\right] \left[\begin{array}{cc}1 & -1\\-3 &0\end{array}\right]

=\left[\begin{array}{cc}6+15& -6+0\\-3+12&3-0\end{array}\right]

Therefore the product of AB is

AB=\left[\begin{array}{cc}21&-6\\ 9&3\end{array}\right]

BA=\left[\begin{array}{cc}1&-1\\-3&0\end{array}\right] \left[\begin{array}{cc}6&-5\\-3&-4\end{array}\right]

=\left[\begin{array}{cc}6+3& -5+4\\-18+0&15+0\end{array}\right]

Therefore the product of BA is

BA=\left[\begin{array}{cc}9&-1\\-18&15\end{array}\right]

The products of AB and BA is given by

AB=\left[\begin{array}{cc}21&-6\\ 9&3\end{array}\right]

BA=\left[\begin{array}{cc}9&-1\\-18&15\end{array}\right]

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Answer:

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