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Slav-nsk [51]
3 years ago
10

a dollhuse has a bed with diemesions 1/12 size of a queen size bed has an area of 4,800 square inches, and a length of 80 inches

. what are the side lengths of the dollhouse bed?
Mathematics
2 answers:
lianna [129]3 years ago
6 0
<span>First we,Find the other side of the bed!

Which would be 60 since 4800/80(the other side) would give us 60

now we take 80 and divide it by 12 ( since the dimension is 1/12) to get 6.667

now we do the same for the 60 ( the other side that we found above)
so 60/12 would equal 5

 So now we multiply the two new found numbers (5 and 6.6667)
 5 * 6 = 30

So your answer would be 30 Inches squared.... I hope that this helps you!!
 I got 20 by 20 equals 400 inches squared so the sides would be 20 message me for more!



</span>
aliina [53]3 years ago
3 0
Divid to get the other side :) so if the doll house is 1/12 of the queen size bed you divid each side by 12 see. 4800/80=60. 80/12=6.6667.... so 60/12=5. so then 5x6 1/3. hope this helps you ... goood luck! :) 
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22.5

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We compute x^2-6x+9 = -3x + 27, thus x^2-6x+9-(-3x + 27) = x^2-3x -18 = 0.

The roots of that quadratic function are

r_1, r_2 = \frac{3 ^+_- \sqrt { 9 +72}}{2} = \frac{3^+_-9}{2} , thus r1 = 6, r2 = -3. We dont care about -3 because it is outside the interval, but we know that f and g graphs intersects on x = 6. Thus, we obtain, due to Bolzano Theorem:

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Hence, the upper bound is f on the interval [3,6) and g on the interval (6,9]. While the lower bound is the 0 function.

We need to calculate the following integral, using Barrow's rule

\int\limits^6_3 {x^2-6x+9} \, dx + \int\limits^9_6 {-3x+27} \, dx = (\frac{x^3}{3} - 3x^2 + 9x) |^6_3 + (\frac{-3x^2}{2} + 27x)|^9_6 = \\  (18 - 9) + (121.5-108) = 22.5

As a result, the area of the region R is 22.5

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