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fgiga [73]
2 years ago
7

How many 2-inch by 3-inch by 8-inch bricks will you need to build a (uniformly wide) brick walk with the shape shown in the figu

re below? (Lay the bricks so that the largest face is up. Assume that a = 10 ft, b = 6 ft, and c = 22 ft.
Mathematics
1 answer:
GalinKa [24]2 years ago
5 0
Here we find the volume of the bricks, will be given by:
volume=length*width*height
v=2×3×8=48 in³

volume of the bricks is given by:
1 ft= 12 inches
thus our dimensions will be:
width=10*12=120 in
height=6*12=72 in
length=22*12=264 in
V=120×72×264
V=2,280,960 in³
Thus the number of bricks required will be:
2280960/48
=47520 in³



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<span>OK, here's a *very* approximate "creative" method: 
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sin(a+b)=sin(a)cos(b) + cos(a)sin(b) </span>
<span>
choose a=30deg and b=5deg </span>
- sin(a)=0.5 (exact) 
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- similarly, for b sufficiently small, cos(b)~1 
<span>- now we only need cos(a) = sqrt(3)/2. We can approximately calculate sqrt(3) as 1.73, (by trial-and-error or better, by using the Babylonian method;  so sqrt(3)/2 = 0.865 </span>
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Finally, sin(30 + 5) = (0.5 x 1) + (0.087 x 0.865) = 0.575 (calculating more digits is of no use because the sin(b)=b and cos(b)=1 are quite inaccurate approximations). </span>

<span>So that wasn't all too much effort, and we're close to the correct answer (0.574)


I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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