Answer:
0 boxes minimum
Step-by-step explanation:
The mass of the truck and paper must satisfy ...
22.5b + 2948.35 ≤ 4700 . . . . total truck mass cannot exceed bridge limits
22.5b ≤ 1751.65
b ≤ 77.85
The driver can take a minimum of 0 boxes and a maximum of 77 boxes of paper over the bridge.
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The question asks for the <em>minimum</em>. We usually expect such a question to ask for the <em>maximum</em>.
You would add the 2 fractions. Since 2/16 equals 1/8 you could add 1/8 plus 4/8. That will give you 5/8. After all that you subtract 5/8 from 8/8, which will give you 3/8.
<h3>
Answer: C) 2.47 cm</h3>
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Explanation:
We have 81 cm^3 of clay. Divide this among the 20 students and each gets 81/20 = 4.05 cm^3 of clay
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The volume of a cone is
V = (1/3)*pi*r^2*h
Solving for h gets us
3V = pi*r^2*h
(3V)/(pi*r^2) = h
h = (3V)/(pi*r^2)
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The diameter is 2.5 cm which cuts in half to 1.25 cm, so this is the radius.
We'll plug this radius in, along with V = 4.05 and pi = 3.14
h = (3V)/(pi*r^2)
h = (3*4.05)/(3.14*(1.25)^2)
h = 2.4764
This value is approximate. Rounding down to the nearest hundredth gets us 2.47
We round down because rounding up to 2.48 will lead to a volume larger than 4.05
Answer:
a
Step-by-step explanation:
just needed the points, sorry
35 + 0.05m = 50 + 0.01m <== ur equation
0.05m - 0.01m = 50 - 35
0.04m = 15
m = 15/0.04
m = 375.....u need to send 375 text messages for the plans to be equal
