Question

the function ( r = 73.3sqrt4{m^3} ), known as kleiber’s law, relates the basal metabolic rate ( r ) in calories per day burned and the body mass ( m ) of a mammal in kilograms. the table shows typical body masses for some members of the cat family.

animal	typical body mass (kg)
cheetah	55
lion	170

i. rewrite the formula using rational exponents.
a. ( r = 73.3(m)^{\frac{1}{4}} )
b. ( r = 73.3(m)^{12} )
c. ( r = 73.3(m)^{\frac{3}{4}} )
d. ( r = (73.3m)^{\frac{1}{4}} )
ii. what is the of ( r ) for a cheetah in terms of calories? round your answer to the nearest whole number.
a. 1,480 calories
b. 15,332 calories
c. 64,164 calories
d. 3,451 calories

you may complete the work on your student work document. make your choices from the dropdown options below.

the formula written with rational exponents is (\boldsymbol{\text{dropdown}}).

the value of ( r ) for a cheetah in terms of calories is (\boldsymbol{\text{dropdown}}).

Explanation:

Step1: Rewrite root as rational exponent

Recall $\sqrt[4]{M^3} = M^{\frac{3}{4}}$, so $R = 73.3M^{\frac{3}{4}}$.

Step2: Identify correct formula match

Compare to options: matches option c.

Step3: Substitute cheetah mass into formula

Cheetah mass $M=55$ kg, so $R = 73.3\times(55)^{\frac{3}{4}}$.

Step4: Calculate $55^{\frac{3}{4}}$ first

$55^{\frac{3}{4}} = (55^{\frac{1}{4}})^3$. $55^{\frac{1}{4}} \approx 2.723$, so $(2.723)^3 \approx 20.23$.

Step5: Compute final R value

$R \approx 73.3 \times 20.23 \approx 1483$, rounds to 1480.

Answer:

i. c. $R = 73.3(M)^{\frac{3}{4}}$
ii. a. 1,480 calories