SOLUTION: Write a proportion for each problem, and solve for the unknown number.
A. Leaf-cutter ants live in central and south america weigh about 1.5grams. One ant can carrry a 4gram pie
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Question 58837This question is from textbook discovering algebra. an investigative approach
: Write a proportion for each problem, and solve for the unknown number.
A. Leaf-cutter ants live in central and south america weigh about 1.5grams. One ant can carrry a 4gram piece of leaf that is about the size of a dime. If a person could carry proportionally as much as the leaf-cutter ant, how much could a 55-kilogram algebra student carry?
B. The leaf cutter ant is about 1.27cm long and takes strides of 0.84cm. If a person could take proportionally equivalent strides, what size strides would a 1.65-m-tall algebra student take?
C. The 1.27-cm-long ants travel up to 0.4 km from home each day. If a person could travel a proportional distance, how far would a 1.65-m-tall person travel?
This question is from textbook discovering algebra. an investigative approach
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Write a proportion for each problem, and solve for the unknown number.
A. Leaf-cutter ants live in central and south america weigh about 1.5grams. One ant can carrry a 4gram piece of leaf that is about the size of a dime. If a person could carry proportionally as much as the leaf-cutter ant, how much could a 55-kilogram algebra student carry?
x/55000 = 4/1.5
x=146666,67 grams (amount the student could carry)
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B. The leaf cutter ant is about 1.27cm long and takes strides of 0.84cm. If a person could take proportionally equivalent strides, what size strides would a 1.65-m-tall algebra student take?
x/165 =0.84/1.27
x=109.13 cm (stride of the algebra student)
C. The 1.27-cm-long ants travel up to 0.4 km from home each day. If a person could travel a proportional distance, how far would a 1.65-m-tall person
travel?
x/400 km = 165/1.27
x=51968.50 cm (distance the 1.65 m tall person would travel)
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Cheers,
Stan H.
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