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1 year ago

Michael read 135 pages in 1 1/2 hours, Select all of the rates that are equivalent to this rate

Mathematics

1 answer:

devlian [24]1 year ago

6 0

The rates that are equivalent to this rate is 90 pages/hr

<h3>Rate of change</h3>

The rate is the number of pages Michael read in 1 hour.

If he read 135 pages in 1 1/2 hours then;

135 pages = 1.5 hours

To determine the rate, we will have:

x = 1 hour

Divide both expressions:

135/x = 1.5/1

1.5x = 135

x = 135/1.5

x = 90

Hence the rates that are equivalent to this rate is 90 pages/hr

Learn more on rate here: brainly.com/question/119866

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❊ Simplify :

DiKsa [7]

Answer:

See Below.

Step-by-step explanation:

Problem 1)

We want to simplify:

$\displaystyle \frac{a+2}{a^2+a-2}+\frac{3}{a^2-1}$

First, let's factor the denominators of each term. For the second term, we can use the difference of two squares. Hence:

$\displaystyle =\frac{a+2}{(a+2)(a-1)}+\frac{3}{(a+1)(a-1)}$

Now, create a common denominator. To do this, we can multiply the first term by (a + 1) and the second term by (a + 2). Hence:

$\displaystyle =\frac{(a+2)(a+1)}{(a+2)(a-1)(a+1)}+\frac{3(a+2)}{(a+2)(a-1)(a+1)}$

Add the fractions:

$\displaystyle =\frac{(a+2)(a+1)+3(a+2)}{(a+2)(a-1)(a+1)}$

Factor:

$\displaystyle =\frac{(a+2)((a+1)+3)}{(a+2)(a-1)(a+1)}$

Simplify:

$\displaystyle =\frac{a+4}{(a-1)(a+1)}$

We can expand. Therefore:

$\displaystyle =\frac{a+4}{a^2-1}$

Problem 2)

We want to simplify:

$\displaystyle \frac{1}{(a-b)(b-c)}+\frac{1}{(c-b)(a-c)}$

Again, let's create a common denominator. First, let's factor out a negative from the second term:

$\displaystyle \begin{aligned} \displaystyle &= \frac{1}{(a-b)(b-c)}+\frac{1}{(-(b-c))(a-c)}\\\\&=\displaystyle \frac{1}{(a-b)(b-c)}-\frac{1}{(b-c)(a-c)}\\\end{aligned}$

Now to create a common denominator, we can multiply the first term by (a - c) and the second term by (a - b). Hence:

$\displaystyle =\frac{(a-c)}{(a-b)(b-c)(a-c)}-\frac{(a-b)}{(a-b)(b-c)(a-c)}$

Subtract the fractions:

$\displaystyle =\frac{(a-c)-(a-b)}{(a-b)(b-c)(a-c)}$

Distribute and simplify:

$\displaystyle =\frac{a-c-a+b}{(a-b)(b-c)(a-c)}=\frac{b-c}{(a-b)(b-c)(a-c)}$

Cancel. Hence:

$\displaystyle =\frac{1}{(a-b)(a-c)}$

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