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Debora [2.8K]
3 years ago
9

A young sumo wrestler decided to go on a special high-protein diet to gain weight rapidly. He gained weight at a rate of 5.55.55

, point, 5 kilograms per month. After 111111 months, he weighed 140140140 kilograms. Let WWW represent the sumo wrestler's weight (in kilograms) after ttt months. Complete the equation for the relationship between the weight and number of months.
Mathematics
1 answer:
Juli2301 [7.4K]3 years ago
7 0

Answer:

The equation for the relationship between the weight and number of months is W(t)= 5.5t+79.5

Step-by-step explanation:

we know that

A linear equation in slope intercept form is equal to

y=mx+b

where

m is the slope

b is the y-coordinate of the y-intercept (value of y when the value of x is equal to zero)

or say b is the actual weight of the wrestler.

In this problem

m=5.5 kg/month

Let

W ⇒ the sumo wrestler's weight in kg (dependent variable)

t ⇒ the time in months (independent variable)

The linear equation is equal to

W(t)=5.5t+b

where

b is the actual weight when he started the diet.

Now Given

W = 140 kg

t = 11 months

substitute in the linear equation to find the value of b

140 = 5.5\times11 +b\\140 = 60.5+b\\b = 140-60.5\\b = 79.5\ kg

W(t)= 5.5t+79.5

Hence,The equation for the relationship between the weight and number of months is W(t)= 5.5t+79.5

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Element X is a radioactive isotope such that its mass decreases by 26% every day. If an experiment starts out with 810 grams of
ioda

Answer:

The hourly decay rate is of 1.25%, so the hourly rate of change is of -1.25%.

The function to represent the mass of the sample after t days is A(t) = 810(0.74)^t

Step-by-step explanation:

Exponential equation of decay:

The exponential equation for the amount of a substance is given by:

A(t) = A(0)(1-r)^t

In which A(0) is the initial amount and r is the decay rate, as a decimal.

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A(t) = A(0)(1-r)^t

0.74A(0) = A(0)(1-r)^{24}

(1-r)^{24} = 0.74

\sqrt[24]{(1-r)^{24}} = \sqrt[24]{0.74}

1 - r = (0.74)^{\frac{1}{24}}

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The hourly decay rate is of 1.25%, so the hourly rate of change is of -1.25%.

Starts out with 810 grams of Element X

This means that A(0) = 810

Element X is a radioactive isotope such that its mass decreases by 26% every day.

This means that we use, for this equation, r = 0.26.

The equation is:

A(t) = A(0)(1-r)^t

A(t) = 810(1 - 0.26)^t

A(t) = 810(0.74)^t

The function to represent the mass of the sample after t days is A(t) = 810(0.74)^t

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