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zalisa [80]
3 years ago
5

A rectangle and a square have the same area. The width of the rectangle is 2 in less than the side of the square and the length

of the rectangle is 3 in less than twice the side of the square. What are the dimensions of the rectangle?
Mathematics
1 answer:
lina2011 [118]3 years ago
5 0
The area of a rectangle is A=LW, the area of a square is A=S^2.

W=S-2 and L=2S-3

And we are told that the areas of each figure are the same.

S^2=LW, using L and W found above we have:

S^2=(2S-3)(S-2)  perform indicated multiplication on right side

S^2=2S^2-4S-3S+6  combine like terms on right side

S^2=2S^2-7S+6  subtract S^2 from both sides

S^2-7S+6=0  factor:

S^2-S-6S+6=0

S(S-1)-6(S-1)=0

(S-6)(S-1)=0, since W=S-2, and W>0, S>2 so:

S=6 is the only valid value for S.  Now we can find the dimensions of the rectangle...

W=S-2 and L=2S-3  given that S=6 in

W=4 in and L=9 in

So the width of the rectangle is 4 inches and the length of the rectangle is 9 inches.


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The question is incomplete. Here is the complete question.

Uninhibited growth can be modeled by exponential functions other than A(t)=A_{0}e^{kt}. for example, if an initial population P₀ requires n units of time to triple, then the function P(t)=P_{0}(3)^{\frac{t}{n} } models the size of the population at time t. An insect population grows exponentially. Complete the parts a through d below.

a) If the population triples in 30 days, and 50 insects are present initially, write an exponential function of the form P(t)=P_{0}(3)^{\frac{t}{n} } that models the population.

b) What will the population be in 47 days?

c) When wil the population reach 750?

d) Express the model from part (a) in the form A(t)=A_{0}e^{kt}.

Answer: a) P(t)=50(3)^{\frac{t}{30} }

              b) P(t) = 280 insects

              c) t = 74 days

             d) A(t)=50e^{0.037t}

Step-by-step explanation:

a) n is time necessary to triple the population of insects, i.e., n = 30 and P₀ = 50. So, Exponential equation for growth is

P(t)=50(3)^{\frac{t}{30} }

b) In t = 47 days:

P(t)=50(3)^{\frac{t}{30} }

P(47)=50(3)^{\frac{47}{30} }

P(47)=50(3)^{1.567}

P(47) = 280

In 47 days, population of insects will be 280

c) P(t) = 750

750=50(3)^{\frac{t}{30} }

\frac{750}{50}=(3)^{\frac{t}{30} }

(3)^{\frac{t}{n} }=15

Using the property <u>Power</u> <u>Rule</u> of logarithm:

log(3)^{\frac{t}{30} }=log15

\frac{t}{30}log(3)=log15

t=\frac{log15}{log3} .30

t = 74

To reach a population of 750 insects, it will take 74 days

d) To express the population growth into the described form, determine the constant k, using the following:

A(t) = 3A₀ and t = 30

A(t)=A_{0}e^{kt}

3A_{0}=A_{0}e^{30k}

3=e^{30k}

Use Power Rule again:

ln3=ln(e^{30k})

ln3=30k

k=\frac{ln3}{30}

k = 0.037

Equation for exponential growth will be:

A(t)=50e^{0.037t}

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