<span>There are 4 long rows and 5 short rows in the theater. x represents the number of chairs in each long row and y represents the number of chairs in each short row.

So, total number of chairs in 4 long rows= 4x
Total number of chairs in 5 short rows = 5y
Total number of chairs in the theater on a normal day = 4x + 5y

When 2 chairs are added to each long row, the number of chairs will change to (x+2).
So, total number of chairs in 4 long rows will be = 4(x+2)

When 3 chairs are added to each short row, the number of chairs will change to (y+3)
So, total number of chairs in 5 short rows will be = 5(y+3)

Thus, total number of chairs in the theater in rush day = 4(x+2) + 5(y+3)
= 4x + 8 + 5y + 15
= 4x + 5y + 23

Thus we can say the number of chairs increase by 23 as compared to a normal day.
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7 0

3 years ago

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Jamal is making two paintings using canvases that are similar rectangles. The length of the smaller canvas is 3 ft and the width

Alex787 [66]

Let's start with what we know:

Smaller canvas:
Length ( $L_{1}$ ) = 3ft
Width ( $W_{1}$ ) = 5ft

Larger canvas:
Length ( $L_{2}$ ) = ?
Width ( $W_{2}$ ) = 10ft

Since these are similar rectangles, we can cross-multiply to calculate the missing length. Here's that formula:

$\frac{ L_{1} }{ L_{2} } = \frac{ W_{1} }{ W_{2} }$
So let's plug it all in from above:

$\frac{ 3 }{ L_{2} } = \frac{ 5 }{ 10 }$
Now we cross multiply by multiplying the top-left by the bottom-right and vice versa:

$(3)(10) = (5)(L_{2})$
$30 = 5L_{2}$
Now divide each side by 5 to isolate $L_{2}$

$\frac{30}{5} = \frac{ 5L_{2}}{5}$
The 5s on the right cancel out, leaving us with:

$6 = L_{2}$

So the length of the larger canvas is 6 ft

4 0

3 years ago

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5 points for the whole problem (Sorry it's all I have left)

Strike441 [17]

Answer: i don't see the problem

Step-by-step explanation:

3 0

3 years ago

Which of the following is most likely the next step in the series<br><br>

poizon [28]

Answer: