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In-s [12.5K]
3 years ago
7

Find the slope of the line containing the points (2, 7) and (-5, -4).

Mathematics
2 answers:
kogti [31]3 years ago
7 0

the answer is 11/7.you can see the image

goldenfox [79]3 years ago
7 0

Answer:

\boxed {\boxed {\sf \frac{11}{7}}}

Step-by-step explanation:

The slope describes the direction and steepness of a line. The formula is:

m= \frac{y_2-y_1}{x_2-x_1}

Where (x₁, y₁) and (x₂, y₂) are the points the line contains. For this problem, the line contains the points (2,7) and (-5, -4). Therefore:

  • x₁= 2
  • y₁ = 7
  • x₂ = -5
  • y₂ = -4

Substitute these values into the formula.

m= \frac{ -4 -7}{-5-2}

Solve the numerator (-4 -7 = -11).

m= \frac{ -11}{-5-2}

Solve the denominator (-5-2 = -7).

m= \frac{ -11}{-7}

Simplify the fraction. The 2 negative signs cancel each other out.

m= \frac{11}{7}

The slope of the line is <u>11/7</u>

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Answer:

38.26 years.

Step-by-step explanation:  

We have been given that the population of an endangered animal by  reduces 8​% per year. the current population of the animal is  1700. When the population of this animal falls below  70​, its extinction is inevitable.  

Let us write the model for population of this animal (y) after x years.

Since we know that an exponential decay function is in form: y=a*(1-r)^x, where,

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8\%=\frac{8}{100}=0.08

Upon substituting a= 1700 and r = 0.08 in exponential function, we will get the model of animal population as:

y=1700*(1-0.08)^x

y=1700*(0.92)^x

Let us find, when the animal population will face extinction by substituting y=70 in our function.

70=1700*(0.92)^x  

\frac{70}{1700}=\frac{1700*(0.92)^x}{1700}

0.0411764705882353=0.92^x  

Let us take natural log of both sides of our equation.

ln(0.0411764705882353)=ln(0.92^x)

ln(0.0411764705882353)=x*ln(0.92)

-3.1898882879949483163=x*-0.0833816089390511

x=\frac{-3.1898882879949483163}{-0.0833816089390511}

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Therefore, the population of the animal will face extinction after 38.26 years.

5 0
4 years ago
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AnnZ [28]

Answer:

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Step-by-step explanation:

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