QuestionSolve the compound inequality $0 \leq s+16 \leq 1$ , graph the solution, and modify one endpoint from closed to open.

Studdy Solution

STEP 1

Assumptions1. The inequality given is $0 \leq s+16 \leq1$ . We are solving for $s$

STEP 2

The given inequality is a compound inequality, which means it is a combination of two inequalities. We can solve this by breaking it up into two separate inequalities and solving each one individually.
The two inequalities are1. $0 \leq s+16$
2. $s+16 \leq1$

STEP 3

Let's solve the first inequality $0 \leq s+16$ .
To do this, we need to isolate $s$ on one side of the inequality. We can do this by subtracting16 from both sides.
$0 -16 \leq s+16 -16$

STEP 4

implify the inequality to get the solution.
$-16 \leq s$

STEP 5

Now, let's solve the second inequality $s+16 \leq1$ .
Again, we need to isolate $s$ on one side of the inequality. We can do this by subtracting16 from both sides.
$s+16 -16 \leq1 -16$

STEP 6

implify the inequality to get the solution.
$s \leq -15$

STEP 7

Now, we have two inequalities1. $-16 \leq s$
2. $s \leq -15$
The solution to the compound inequality is the intersection of the solutions to these two inequalities, which is $-16 \leq s \leq -15$ .

STEP 8

To graph the solution, we plot the interval $[-16, -15]$ on the number line. The endpoints -16 and -15 are included in the solution, so we use closed circles at these points.

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QuestionSolve the compound inequality $0 \leq s+16 \leq 1$ , graph the solution, and modify one endpoint from closed to open.

Studdy Solution

STEP 1

Assumptions1. The inequality given is $0 \leq s+16 \leq1$ . We are solving for $s$

STEP 2

The given inequality is a compound inequality, which means it is a combination of two inequalities. We can solve this by breaking it up into two separate inequalities and solving each one individually.
The two inequalities are1. $0 \leq s+16$
2. $s+16 \leq1$

STEP 3

Let's solve the first inequality $0 \leq s+16$ .
To do this, we need to isolate $s$ on one side of the inequality. We can do this by subtracting16 from both sides.
$0 -16 \leq s+16 -16$

STEP 4

implify the inequality to get the solution.
$-16 \leq s$

STEP 5

Now, let's solve the second inequality $s+16 \leq1$ .
Again, we need to isolate $s$ on one side of the inequality. We can do this by subtracting16 from both sides.
$s+16 -16 \leq1 -16$

STEP 6

implify the inequality to get the solution.
$s \leq -15$

STEP 7

Now, we have two inequalities1. $-16 \leq s$
2. $s \leq -15$
The solution to the compound inequality is the intersection of the solutions to these two inequalities, which is $-16 \leq s \leq -15$ .

STEP 8

To graph the solution, we plot the interval $[-16, -15]$ on the number line. The endpoints -16 and -15 are included in the solution, so we use closed circles at these points.

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QuestionSolve the compound inequality 0≤s+16≤10 \leq s+16 \leq 10≤s+16≤1, graph the solution, and modify one endpoint from closed to open.

Studdy Solution

STEP 1

Assumptions1. The inequality given is 0≤s+16≤10 \leq s+16 \leq10≤s+16≤1 . We are solving for sss

STEP 2

The given inequality is a compound inequality, which means it is a combination of two inequalities. We can solve this by breaking it up into two separate inequalities and solving each one individually.The two inequalities are1. 0≤s+160 \leq s+160≤s+162. s+16≤1s+16 \leq1s+16≤1

STEP 3

Let's solve the first inequality 0≤s+160 \leq s+160≤s+16.To do this, we need to isolate sss on one side of the inequality. We can do this by subtracting16 from both sides.0−16≤s+16−160 -16 \leq s+16 -160−16≤s+16−16

STEP 4

implify the inequality to get the solution.−16≤s-16 \leq s−16≤s

STEP 5

Now, let's solve the second inequality s+16≤1s+16 \leq1s+16≤1.Again, we need to isolate sss on one side of the inequality. We can do this by subtracting16 from both sides.s+16−16≤1−16s+16 -16 \leq1 -16s+16−16≤1−16

STEP 6

implify the inequality to get the solution.s≤−15s \leq -15s≤−15

STEP 7

Now, we have two inequalities1. −16≤s-16 \leq s−16≤s2. s≤−15s \leq -15s≤−15The solution to the compound inequality is the intersection of the solutions to these two inequalities, which is −16≤s≤−15-16 \leq s \leq -15−16≤s≤−15.

STEP 8

To graph the solution, we plot the interval [−16,−15][-16, -15][−16,−15] on the number line. The endpoints -16 and -15 are included in the solution, so we use closed circles at these points.

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Studdy solves anything!

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QuestionSolve the compound inequality $0 \leq s+16 \leq 1$ , graph the solution, and modify one endpoint from closed to open.

Assumptions1. The inequality given is $0 \leq s+16 \leq1$ . We are solving for $s$

The given inequality is a compound inequality, which means it is a combination of two inequalities. We can solve this by breaking it up into two separate inequalities and solving each one individually.
The two inequalities are1. $0 \leq s+16$
2. $s+16 \leq1$

Let's solve the first inequality $0 \leq s+16$ .
To do this, we need to isolate $s$ on one side of the inequality. We can do this by subtracting16 from both sides.
$0 -16 \leq s+16 -16$

implify the inequality to get the solution.
$-16 \leq s$

Now, let's solve the second inequality $s+16 \leq1$ .
Again, we need to isolate $s$ on one side of the inequality. We can do this by subtracting16 from both sides.
$s+16 -16 \leq1 -16$

implify the inequality to get the solution.
$s \leq -15$

Now, we have two inequalities1. $-16 \leq s$
2. $s \leq -15$
The solution to the compound inequality is the intersection of the solutions to these two inequalities, which is $-16 \leq s \leq -15$ .

To graph the solution, we plot the interval $[-16, -15]$ on the number line. The endpoints -16 and -15 are included in the solution, so we use closed circles at these points.