The equation that models the exponential decline is y = 46348587109809610(0.9861)ˣ if the population over the past 20 years is provided.

A mathematical notion called the line of the best fit connects points spread throughout a graph. It's a type of linear regression that uses scatter data to figure out the best way to define the dots' relationship.

Let's suppose the equation that models the exponential decline is:

$\rm y = a(b^x)$

From the data given, we can calculate the value of a and b:

a = 46348587109809610

b = 0.9861

y = 46348587109809610(0.9861)ˣ

Thus, the equation that models the exponential decline is y = 46348587109809610(0.9861)ˣ if the population over the past 20 years is provided.

Learn more about the line of best fit here:

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6 0

2 years ago

What is a solution to the equation 3 / m + 3 - M / 3 - M equals m^2 + 9 / m^2-9?

Mnenie [13.5K]

Answer: Last option.

Step-by-step explanation:

Given the equation:

$\frac{3}{m+3}-\frac{m}{3-m}=\frac{m^2+9}{m^2-9}$

Follow these steps to solve it:

- Subtract the fractions on the left side of the equation:

$\frac{3(3-m)-m(m+3)}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{9-3m-m^2-3m}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}\\\\\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{m^2-9}$

- Using the Difference of squares formula ( $a^2-b^2=(a+b)(a-b)$ ) we can simplify the denominator of the right side of the equation:

$\frac{-m^2-6m+9}{(m+3)(3-m)}=\frac{m^2+9}{(m+3)(m-3)}$

- Multiply both sides of the equation by $(m+3)(3-m)$ and simplify:

$\frac{(-m^2-6m+9)(m+3)(3-m)}{(m+3)(3-m)}=\frac{(m^2+9)(m+3)(3-m)}{(m+3)(m-3)}\\\\-m^2-6m+9=\frac{(m^2+9)(3-m)}{(m-3)}$

- Multiply both sides by $m-3$ :

$(-m^2-6m+9)(m-3)=\frac{(m^2+9)(3-m)(m-3)}{(m-3)}\\\\(-m^2-6m+9)(m-3)=(m^2+9)(3-m)$

- Apply Distributive property and simplify:

$(-m^2-6m+9)(m-3)=(m^2+9)(3-m)\\\\-m^3-6m^2+9m+3m^2+18m-27=3m^2+27-m^3-9m\\\\-m^3-3m^2+27m-27+m^3-3m^2+9m-27=0\\\\-6m^2+36m-54=0$

- Divide both sides of the equation by -6:

$\frac{-6m^2+36m-54}{-6}=\frac{0}{-6}\\\\m^2-6m+9=0$

- Factor the equation and solve for "m":

$(m-3)^2=0\\\\m=3$

In order to verify it, you must substitute $m=3$ into the equation and solve it:

$\frac{3}{3+3}-\frac{3}{3-3}=\frac{3^2+9}{3^2-9}\\\\\frac{3}{6}-\frac{3}{0}=\frac{18}{0}$

<em>NO SOLUTION</em>

7 0

3 years ago