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3 years ago

Solving Absolute Value Equations: Solve 4|x+7|+8=32

Mathematics

1 answer:

Len [333]3 years ago

3 0

Here’s if you need it written out

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Round 0.978 to the nearest tenth

Setler79 [48]

The answer is one, because the number seven in the hundredths column is greater than five so you round the nine up to 1.0.

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3 years ago

Why does math even exist?

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3 years ago

Read 2 more answers

A cookout tray consists of one main item, two different sides, and a drink. if cookout offers 10 main items, 10 sides, and 5 dri

OLEGan [10]

The number of unique cookout trays are possible is 500

<h3>How many unique cookout trays are possible?</h3>

The given parameters are:

Main items = 10

Sides = 10

Drinks = 5

The number of unique cookout trays are possible is

Cookout trays = Main items * Sides * Drinks

So, we have:

Cookout trays = 10 * 10 * 5

Evaluate

Cookout trays = 500

Hence, the number of unique cookout trays are possible is 500

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brainly.com/question/11732255

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1 year ago

Submission in an houuuur!!! Please send help and can I please have the explanation.

Ksenya-84 [330]

Answer:

It should 3/10

Step-by-step explanation:

3 0

3 years ago

❊ 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:

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

4 0

3 years ago