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Solve for $r$ . $\newline$ $|r| + 1 \leq 13$ $\newline$ Write a compound inequality like $1 < x < 3$ or like $x < 1$ or $x > 3$ . Use integers, proper fractions, or improper fractions in simplest form. $\newline$ ______

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Solve absolute value inequalities

Full solution

Q. Solve for

r

.

\newline

|r| + 1 \leq 13

\newline

Write a compound inequality like

1 < x < 3

or like

x < 1

or

x > 3

. Use integers, proper fractions, or improper fractions in simplest form.

\newline

______

Isolate absolute value expression: First, we need to isolate the absolute value expression on one side of the inequality. $\newline$ $|r| + 1 \leq 13$ $\newline$ Subtract $1$ from both sides to isolate $|r|$ . $\newline$ $|r| + 1 - 1 \leq 13 - 1$ $\newline$ $|r| \leq 12$
Consider absolute value definition: Now, we need to consider the definition of absolute value, which states that $|r| ) is the distance of \$r$ from $0$ on the number line. This means that $r$ can be either positive or negative, but not greater than $12$ in magnitude. $\newline$ So we have two cases: $\newline$ $1$ . $r \leq 12$ (when $r$ is positive or zero) $\newline$ $2$ . $-r \leq 12$ (when $r$ is negative)
Solve for negative values: For the second case, we need to solve for $r$ . To do this, we multiply both sides by $-1$ , remembering to reverse the inequality sign because we are multiplying by a negative number. $\newline$ $-r \leq 12$ $\newline$ $(-1)(-r) \geq (-1)(12)$ $\newline$ $r \geq -12$
Combine both cases: Combining both cases, we get the compound inequality that represents all possible values of $r$ . $-12 \leq r \leq 12$ This inequality shows that $r$ can be any number between $-12$ and $12$ , inclusive.

More problems from Solve absolute value inequalities

Question

u

\newline

|u| - 4 \geq 6

\newline

\newline

Write a compound inequality like

1 < x < 3

x < 1

x > 3

. Use integers, proper fractions, or improper fractions in simplest form.

\newline

Posted 2 months ago

Question

9 \frac{7}{8}

into an improper fraction.

\newline

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Question

f(x)=\left\{\begin{array}{lll} -4 x & \text { for } & x<-1 \\ -x+1 & \text { for } & -1 \leq x \leq 3 \\ -2 x+8 & \text { for } & x>3 \end{array}\right.

\newline

f(3)

\newline

\newline

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Question

\newline

Solve the system of equations.

\newline

\begin{array}{l} y=22 x-38 \\ y=x^{2}+11 x-14 \end{array}

\newline

Write the coordinates in exact form. Simplify all fractions and radicals.

\newline

\begin{array}{l} (-,-) \\ (-,-) \end{array}

\newline

\rightarrow

Posted 2 months ago

Question

A baseball player went up to bat

250

times in a season. He hit the ball

100

times. Find the rate of balls hit to times at bat. Express as a simplified fraction.

\newline

\frac{2}{5}

\newline

\frac{5}{2}

\newline

\frac{3}{5}

\newline

\frac{2}{3}

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Question

1,2,3,4,5

6

are randomly arranged to form a three-digit number. (Digits are not repeated.) Find the probability that the number is even and greater than

600

\newline

The probability that the three-digit number is even and greater than

600

\square

\newline

(Type an integer or a simplified fraction.)

Posted 2 months ago

Question

Use benchmark fractions to estimate sums and differences less than or gremet than

1

. Write each expresston in the correct answer space.

\newline

\begin{array}{l} \frac{7}{8}+\frac{5}{10} \\ 1 \frac{5}{8}-\frac{5}{6} \\ \frac{10}{10}-\frac{2}{3} \\ \frac{1}{2}+\frac{2}{3} \\ \frac{5}{12}+\frac{1}{4} \\ 1 \frac{1}{6}+\frac{7}{8} \\ \end{array}

Posted 2 months ago

Question

Eric is sorting pencils into boxes. He put

16

pencils in the first box,

21

pencils in the second box,

26

pencils in the third box, and

31

pencils in the fourth box. What kind of sequence is this?

\newline

\newline

\newline

\newline

\newline

Posted 1 month ago

Question

Given the definitions of

f(x)

g(x)

below, find the value of

g(f(2))

\newline

\begin{array}{l} f(x)=-5 x+3 \\ g(x)=2 x^{2}+5 x-14 \end{array}

\newline

Posted 2 months ago

Question

t

\newline

|t + 8| \leq 4

\newline

Write a compound inequality like

1 < x < 3

x < 1

x > 3

. Use integers, proper fractions, or improper fractions in simplest form.

\newline

Posted 1 month ago