Help me i need it

Morgan is throwing a pizza party. She has 1 7/8 pounds of cheese available and she needs 1/5 pound of cheese to make each pizz

Karolina [17]

We have been given that Morgan is throwing a pizza party. She has 1 7/8 pounds of cheese available and she needs 1/5 pound of cheese to make each pizza. We are asked to find the number of full pizzas that Morgan can make with the amount of cheese she has.

To solve our given problem, we will divide total amount of cheese by amount of cheese used to make each pizza.

$\text{Number of pizzas}=1\frac{7}{8}\div \frac{1}{5}$

$\text{Number of pizzas}=\frac{15}{8}\div \frac{1}{5}$

Now we will convert our division problem into multiplication problem by flipping the 2nd fraction.

$\text{Number of pizzas}=\frac{15}{8}\times \frac{5}{1}$

$\text{Number of pizzas}=\frac{75}{8}$

$\text{Number of pizzas}=9\frac{3}{8}$

We are asked to find the number of full pizzas and we can see that Morgan can make 9 full and $\frac{3}{8}$ pizzas.

Since there are 9 full pizzas, therefore, Morgan can make 9 full pizzas with the amount of cheese she has.

3 0

3 years ago

How do you solve these equations? I don't want you to answer all of them, just tell me how to solve each type of equation on the

miss Akunina [59]

Answer:

8. Identify the common denominator; express each fraction using that denominator; combine the numerators of those rewritten fractions and express the result over the common denominator. Factor out any common factors from numerator and denominator in your result. (It's exactly the same set of instructions that apply for completely numerical fractions.)

9. As with numerical fractions, multiply the numerator by the inverse of the denominator; cancel common factors from numerator and denominator.

10. The method often recommended is to multiply the equation by a common denominator to eliminate the fractions. Then solve in the usual way. Check all answers. If one of the answers makes your multiplier (common denominator) be zero, it is extraneous. (10a cannot have extraneous solutions; 10b might)

Step-by-step explanation:

For a couple of these, it is helpful to remember that (a-b) = -(b-a).

$\dfrac{5}{x+2}+\dfrac{25-x}{x^2-3x-10}=\dfrac{5(x-5)}{(x+2)(x-5)}+\dfrac{25-x}{(x+2)(x-5)}\\\\=\dfrac{5x-25+25-x}{(x+2)(x-5)}=\dfrac{4x}{x^2-3x-10}$

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$\displaystyle\frac{\left(\frac{x}{x-2}\right)}{\left(\frac{2x}{2-x}\right)}=\frac{x}{x-2}\cdot\frac{-(x-2)}{2x}=\frac{-x(x-2)}{2x(x-2)}=-\frac{1}{2}$

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$\dfrac{3}{x-1}+\dfrac{6}{x^2-3x+2}=2\\\\\dfrac{3(x-2)}{(x-1)(x-2)}+\dfrac{6}{(x-1)(x-2)}=\dfrac{2(x-1)(x-2)}{(x-1)(x-2)}\\\\3x-6+6=2(x^2-3x+2) \qquad\text{multiply by the denominator}\\\\2x^2-9x+4=0 \qquad\text{subtract 3x}\\\\(2x-1)(x-4)=0 \qquad\text{factor; x=1/2, x=4}$