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marusya05 [52]
3 years ago
10

What is the range of possible sizes for side x? _ < x < _

Mathematics
2 answers:
qwelly [4]3 years ago
8 0

Answer:

2.8 < x < 5.8

Step-by-step explanation:

We must apply the Triangle Inequality Theorem which states that for any triangle with sides a, b, and c:

a + b > c

b + c > a

c + a > b

Here, let's arbitrarily denote a as 4.1, b as 1.3, and c as x. So, let's plug these values into the 3 inequalities listed above:

a + b > c  ⇒  4.1 + 1.3 > x  ⇒  5.8 > x

b + c > a  ⇒  1.3 + x > 4.1  ⇒  x > 2.8

c + a > b  ⇒  x + 4.1 > 1.3  ⇒  x > -2.8

Look at the last two: clearly if x is greater than 2.8 (from the second one), then it will definitely be greater than -2.8 (from the third), so we can just disregard the last inequality.

Thus, the range of possible sizes for x are:

2.8 < x < 5.8

<em>~ an aesthetics lover</em>

zysi [14]3 years ago
8 0

Answer:

There are two cases: either x is the largest value or 4.1 is the largest value.

First case: x is largest so that means x < 4.1 + 1.3 which becomes x < 5.8

Second case: 4.1 is largest so that means x + 1.3 > 4.1 which becomes x > 2.8

Answer is 2.8 < x < 5.8.

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The fox population in a certain region has a continuous growth rate of 5% per year. It is estimated that the population in the y
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Answer:

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Step-by-step explanation:

Given the following :

Continuous growth rate(r) = 5% = 0.05

Population in year 2000 = Initial population (A) = 10,100

Time(t) = period (years since year 2000)

A)

Find a function that models the population,P(t) , after (t) years since year 2000 (i.e. t= 0 for the year 2000).

P(t) = A * (1 + r)^t

Trying out our function for t = year 2000, t =0

P(0) = 10,100 * (1 + 0.05)^0

P(0) = 10,100 * 1.05^0 = 10,100

B.)

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Year 2008, t = 8

P(8) = 10,100 * (1 + 0.05)^8

P(8) = 10,100 * 1. 05^8

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= 14922.29

= 14,922

c) Use your function to estimate the year when the fox population will reach over 18,400 foxes. Round t to the nearest whole year, then state the year.

P(t) = A * (1 + r)^t

18400 = 10,100 * (1.05)^t

18400/10100 = 1.05^t

1.8217821 = 1.05^t

1.05^t = 1.8217821

In(1.05^t) = ln(1.8217821)

0.0487901 * t = 0.5998151

t = 0.5998151 / 0.0487901

t = 12.293787

Therefore eit will take 13 years

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