Math / Algebra

Question43. [-/0.31 Points] DETAILS MY NOTES SCOLALG7 2.2.083. O/100 Submissions Used ASK YOUR TEACHER your answer in dollars.) $\left\{\begin{array}{ll} \square & \text { if } 0<x \leq 1 \\ \square & \text { if } 1<x \leq 2 \\ \text { if } 2<x \leq 3 \\ & \text { if } 3<x \leq 3.5 \end{array}\right.$

Studdy Solution

STEP 1

What is this asking? We need to write a formula for the cost of mailing a letter weighing up to 3.5 ounces, where the first ounce costs $\$0.49$ and each additional ounce (or part of one) costs $\$0.21$ . Watch out! Even if a letter weighs slightly more than a whole ounce, we have to pay for the full extra ounce!

STEP 2

1. Cost for the first ounce
2. Cost for up to two ounces
3. Cost for up to three ounces
4. Cost for up to three and a half ounces

STEP 3

If our letter weighs between zero and one ounce, the cost is simply the initial cost of $\$0.49$ .
So, for $0 < x \le 1$ , the cost is $\$0.49$ .

STEP 4

If our letter weighs between one and two ounces, we pay the base cost of $\$0.49$ plus one additional ounce, which costs $\$0.21$ .

STEP 5

So, we add those together: $\$0.49 + \$0.21 = \$0.70$ .
Therefore, for $1 < x \le 2$ , the cost is $\$0.70$ .

STEP 6

Now, if our letter weighs between two and three ounces, we pay the base cost of $\$0.49$ plus $\$0.21$ for two additional ounces.

STEP 7

Two additional ounces cost $2 \cdot \$0.21 = \$0.42$ .
Adding this to our base cost: $\$0.49 + \$0.42 = \$0.91$ .
So, for $2 < x \le 3$ , the cost is $\$0.91$ .

STEP 8

Finally, if our letter weighs between three and three and a half ounces, we pay the base cost of $\$0.49$ plus $\$0.21$ for three additional ounces (even though it's not quite a full three ounces, we still round up!).

STEP 9

Three additional ounces cost $3 \cdot \$0.21 = \$0.63$ .
Adding this to our base cost: $\$0.49 + \$0.63 = \$1.12$ .
So, for $3 < x \le 3.5$ , the cost is $\$1.12$ .

STEP 10

$\text{Cost}(x) = \begin{cases} \$0.49, & 0 < x \le 1 \\ \$0.70, & 1 < x \le 2 \\ \$0.91, & 2 < x \le 3 \\ \$1.12, & 3 < x \le 3.5 \end{cases}$

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Studdy Solution

STEP 1

STEP 2

1. Cost for the first ounce
2. Cost for up to two ounces
3. Cost for up to three ounces
4. Cost for up to three and a half ounces

STEP 3

If our letter weighs between zero and one ounce, the cost is simply the initial cost of $\$0.49$ .
So, for $0 < x \le 1$ , the cost is $\$0.49$ .

STEP 4

If our letter weighs between one and two ounces, we pay the base cost of $\$0.49$ plus one additional ounce, which costs $\$0.21$ .

STEP 5

So, we add those together: $\$0.49 + \$0.21 = \$0.70$ .
Therefore, for $1 < x \le 2$ , the cost is $\$0.70$ .

STEP 6

Now, if our letter weighs between two and three ounces, we pay the base cost of $\$0.49$ plus $\$0.21$ for two additional ounces.

STEP 7

Two additional ounces cost $2 \cdot \$0.21 = \$0.42$ .
Adding this to our base cost: $\$0.49 + \$0.42 = \$0.91$ .
So, for $2 < x \le 3$ , the cost is $\$0.91$ .

STEP 8

Finally, if our letter weighs between three and three and a half ounces, we pay the base cost of $\$0.49$ plus $\$0.21$ for three additional ounces (even though it's not quite a full three ounces, we still round up!).

STEP 9

Three additional ounces cost $3 \cdot \$0.21 = \$0.63$ .
Adding this to our base cost: $\$0.49 + \$0.63 = \$1.12$ .
So, for $3 < x \le 3.5$ , the cost is $\$1.12$ .

STEP 10

$\text{Cost}(x) = \begin{cases} \$0.49, & 0 < x \le 1 \\ \$0.70, & 1 < x \le 2 \\ \$0.91, & 2 < x \le 3 \\ \$1.12, & 3 < x \le 3.5 \end{cases}$

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Studdy Solution

STEP 1

STEP 2

1. Cost for the first ounce2. Cost for up to two ounces3. Cost for up to three ounces4. Cost for up to three and a half ounces

STEP 3

If our letter weighs between zero and one ounce, the cost is simply the **initial cost** of $0.49\$0.49$0.49.So, for 0<x≤10 < x \le 10<x≤1, the cost is $0.49\$0.49$0.49.

STEP 4

If our letter weighs between one and two ounces, we pay the **base cost** of $0.49\$0.49$0.49 *plus* one additional ounce, which costs $0.21\$0.21$0.21.

STEP 5

So, we **add** those together: $0.49+$0.21=$0.70\$0.49 + \$0.21 = \$0.70$0.49+$0.21=$0.70.Therefore, for 1<x≤21 < x \le 21<x≤2, the cost is $0.70\$0.70$0.70.

STEP 6

Now, if our letter weighs between two and three ounces, we pay the **base cost** of $0.49\$0.49$0.49 *plus* $0.21\$0.21$0.21 for *two* additional ounces.

STEP 7

Two additional ounces cost 2⋅$0.21=$0.422 \cdot \$0.21 = \$0.422⋅$0.21=$0.42.Adding this to our **base cost**: $0.49+$0.42=$0.91\$0.49 + \$0.42 = \$0.91$0.49+$0.42=$0.91.So, for 2<x≤32 < x \le 32<x≤3, the cost is $0.91\$0.91$0.91.

STEP 8

Finally, if our letter weighs between three and three and a half ounces, we pay the **base cost** of $0.49\$0.49$0.49 *plus* $0.21\$0.21$0.21 for *three* additional ounces (even though it's not quite a full three ounces, we still round up!).

STEP 9

Three additional ounces cost 3⋅$0.21=$0.633 \cdot \$0.21 = \$0.633⋅$0.21=$0.63.Adding this to our **base cost**: $0.49+$0.63=$1.12\$0.49 + \$0.63 = \$1.12$0.49+$0.63=$1.12.So, for 3<x≤3.53 < x \le 3.53<x≤3.5, the cost is $1.12\$1.12$1.12.

STEP 10

Cost(x)={$0.49,0<x≤1$0.70,1<x≤2$0.91,2<x≤3$1.12,3<x≤3.5\text{Cost}(x) = \begin{cases} \$0.49, & 0 < x \le 1 \\ \$0.70, & 1 < x \le 2 \\ \$0.91, & 2 < x \le 3 \\ \$1.12, & 3 < x \le 3.5 \end{cases}Cost(x)=⎩⎨⎧​$0.49,$0.70,$0.91,$1.12,​0<x≤11<x≤22<x≤33<x≤3.5​

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1. Cost for the first ounce
2. Cost for up to two ounces
3. Cost for up to three ounces
4. Cost for up to three and a half ounces

If our letter weighs between zero and one ounce, the cost is simply the initial cost of $\$0.49$ .
So, for $0 < x \le 1$ , the cost is $\$0.49$ .

If our letter weighs between one and two ounces, we pay the base cost of $\$0.49$ plus one additional ounce, which costs $\$0.21$ .

So, we add those together: $\$0.49 + \$0.21 = \$0.70$ .
Therefore, for $1 < x \le 2$ , the cost is $\$0.70$ .

Now, if our letter weighs between two and three ounces, we pay the base cost of $\$0.49$ plus $\$0.21$ for two additional ounces.

Two additional ounces cost $2 \cdot \$0.21 = \$0.42$ .
Adding this to our base cost: $\$0.49 + \$0.42 = \$0.91$ .
So, for $2 < x \le 3$ , the cost is $\$0.91$ .

Finally, if our letter weighs between three and three and a half ounces, we pay the base cost of $\$0.49$ plus $\$0.21$ for three additional ounces (even though it's not quite a full three ounces, we still round up!).

Three additional ounces cost $3 \cdot \$0.21 = \$0.63$ .
Adding this to our base cost: $\$0.49 + \$0.63 = \$1.12$ .
So, for $3 < x \le 3.5$ , the cost is $\$1.12$ .

$\text{Cost}(x) = \begin{cases} \$0.49, & 0 < x \le 1 \\ \$0.70, & 1 < x \le 2 \\ \$0.91, & 2 < x \le 3 \\ \$1.12, & 3 < x \le 3.5 \end{cases}$