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Irina-Kira [14]
3 years ago
9

The cost to mall a package is $5 for the first 2 pounds and 40 cents for each additional ounce. Which of the following functions

represents the cost to mall a package if x is the number of ounces over 2 pounds?
plz hurry
​
Mathematics
1 answer:
erica [24]3 years ago
5 0

Answer:

Step-by-step explanation:

5 times 2 equales 10 divided by 40 eqauls 4 times 2 equals 8

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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
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