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zzz [600]
3 years ago
15

Round 75 to the nearest ten.

Mathematics
1 answer:
zhenek [66]3 years ago
8 0

Answer:

75>80

92>90

80×90=7,200.

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14<br> 10<br> 56<br> 40<br> 14<br> x<br> 56<br> find the length of the missing side
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Step-by-step explanation:

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3 years ago
A rectangular enclosure is to be created using 82m rope.
atroni [7]
Let
x-----> the length of rectangle
y-----> the width of rectangle

we know that 
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perimeter of rectangle=82 m
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Area of rectangle=x*y
substitute equation 1 in the area formula
Area=x*[41-x]----> 41x-x²

using a graph tool
see the attached figure

the vertex is the point (20.5,420.25)
that means
 for x=20.5 m ( length of rectangle)
the area is 420.25 m²

y=420.25/20.5----> 20.5 m

the dimensions are
20.5 m x 20.5 m------> is a square

the answer part 1) 
<span>the dimensions of the rectangular with Maximum area is a square with length side 20.5 meters
</span>
Part 2)<span>b) Suppose 41 barriers each 2m long, are used instead. Can the same area be enclosed?
</span>divide the length side of the square by 2
so
20.5/2=10.25--------> 10 barriers
the dimensions are 10 barriers x 10 barriers
10 barriers=10*2---> 20 m

the area enclosed with barriers is =20*20----> 400 m²

400 m² < 420.25 m²
so

the answer Part 2) is 
<span>the area enclosed by the barriers is less than the area enclosed by the rope
</span>
Part 3)<span>How much more area can be enclosed if the rope is used instead of the barriers
</span>
area using the rope=420.25 m²
area using the barriers=400 m²

420.25-400=20.25 m²

the answer part 3) is
20.25 m²

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3 years ago
Consider the Fourier sine series of each of the following functions. In this exercise do not compute the coefficients but use th
tekilochka [14]

Answer:

Step-by-step explanation:

6 0
3 years ago
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