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Sagot :
The rectangle problem:
First, create some variables.
w = width
l = length
Then you know that the length is 3 times the width, so
l = 3w
The equation for perimeter is
P = 2(l+w)
Now you use substitution, substituting the 3w for l because you found that earlier.
P = 2((3w) + w)
And simplify.
P = 2(4w)
P = 8w
And you know P = 30 because they told you that.
30 = 8w
w = 15/4 inches, or 3.75 inches
Now substitute that value back into that first equation,
l = 3w
l = 3(15/4)
l = 45/4 inches, or 11.25 inches.
The dimensions are 11.25in for length and 3.75in for width.
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The volleyball problem:
100 minutes of stretching and scrimmaging combined, and you know 10 have to be stretching, so 90 is scrimmaging.
s = scrimmaging
T = total time
s = 2/3T (they gave you this)
90 = 2/3T
T = (90*3)/2
T = 270/2
T = 135 minutes, or 2 hours and 15 minutes.
First, create some variables.
w = width
l = length
Then you know that the length is 3 times the width, so
l = 3w
The equation for perimeter is
P = 2(l+w)
Now you use substitution, substituting the 3w for l because you found that earlier.
P = 2((3w) + w)
And simplify.
P = 2(4w)
P = 8w
And you know P = 30 because they told you that.
30 = 8w
w = 15/4 inches, or 3.75 inches
Now substitute that value back into that first equation,
l = 3w
l = 3(15/4)
l = 45/4 inches, or 11.25 inches.
The dimensions are 11.25in for length and 3.75in for width.
----------------------------------------------------------------------------
The volleyball problem:
100 minutes of stretching and scrimmaging combined, and you know 10 have to be stretching, so 90 is scrimmaging.
s = scrimmaging
T = total time
s = 2/3T (they gave you this)
90 = 2/3T
T = (90*3)/2
T = 270/2
T = 135 minutes, or 2 hours and 15 minutes.
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