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PROBLEM

have elapsed we use the first formula. If more than 3 hours and up to 5 hours have elapsed, we use the second formula.
Use the formula below to determine each of the following values. Pay attention to the value of to make sure you are using the correct formula!
d={2t if 0≤t≤34(t−3)+6 if 3<t≤5d=\left\{\begin{array}{ccc} 2 t & \text { if } & 0 \leq t \leq 3 \\ 4(t-3)+6 & \text { if } & 3<t \leq 5 \end{array}\right. a. If t=1.75t=1.75, then the total accumulated snowfall up to that point in the storm was □\square inches.
b. If t=4.25t=4.25, then the total accumulated snowfall up to that point in the storm was □\square inches.
c. If t=2.9t=2.9, then the total accumulated snowfall up to that point in the storm was □\square inches.
d. If t=3.1t=3.1, then the total accumulated snowfall up to that point in the storm was □\square inches.
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menes per hour for 1.5 hours (4.5−3=1.5)(4.5-3=1.5). At 5:00pm5: 00 \mathrm{pm}, th has only been falling at a rate of 4 inches per hour for 2 hours 5−3=25-3=2 ).
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d=4(t−3)+6d=4(t-3)+6 Putting everything together we have so far, we get the following. This notation indicates that dd us represented differently dependin on how much time has passed.
d={2t if 0≤t≤34(t−3)+6 if 3<t≤5d=\left\{\begin{array}{ccc} 2 t & \text { if } & 0 \leq t \leq 3 \\ 4(t-3)+6 & \text { if } & 3<t \leq 5 \end{array}\right. If between 0 and 3 hours nave elapsed we use the first formula. If more than 3 hours and up to 5 hours have elapsed, we use the sec formula.
Use the formula below to determine each of the following values. Pay attention to the value of to make sure you are using the correct formula!

STEP 1

What is this asking?
Given a formula for snowfall accumulation that changes depending on how much time has passed, we need to calculate the total snowfall at different times.
Watch out!
We need to make sure we use the right formula based on the time t, and not mix them up!

STEP 2

1. Snowfall at t=1.75t = 1.75
2. Snowfall at t=4.25t = 4.25
3. Snowfall at t=2.9t = 2.9
4. Snowfall at t=3.1t = 3.1

STEP 3

Since our time t=1.75t = 1.75 is between 0 and 3, we'll use the first formula: d=2⋅td = 2 \cdot t.
It's the right one because 0≤1.75≤30 \leq 1.75 \leq 3 is a true statement!

STEP 4

Substitute the value of tt:
d=2⋅1.75d = 2 \cdot 1.75

STEP 5

Calculate the result:
d=3.5d = 3.5

STEP 6

This time, t=4.25t = 4.25 is between 3 and 5, so we're going with the second formula: d=4(t−3)+6d = 4(t - 3) + 6.
This is because 3<4.25≤53 < 4.25 \leq 5 is true!

STEP 7

Substitute t=4.25t = 4.25:
d=4(4.25−3)+6d = 4(4.25 - 3) + 6

STEP 8

Simplify inside the parentheses:
d=4(1.25)+6d = 4(1.25) + 6

STEP 9

Multiply:
d=5+6d = 5 + 6

STEP 10

Add to find our answer:
d=11d = 11

STEP 11

With t=2.9t = 2.9, we're back to the first formula d=2⋅td = 2 \cdot t since 0≤2.9≤30 \leq 2.9 \leq 3.

STEP 12

Substitute t=2.9t = 2.9:
d=2⋅2.9d = 2 \cdot 2.9

STEP 13

Calculate:
d=5.8d = 5.8

STEP 14

Here, t=3.1t = 3.1 falls between 3 and 5, so we'll use d=4(t−3)+6d = 4(t - 3) + 6 because 3<3.1≤53 < 3.1 \leq 5 is true!

STEP 15

Substitute t=3.1t = 3.1:
d=4(3.1−3)+6d = 4(3.1 - 3) + 6

STEP 16

Simplify the parentheses:
d=4(0.1)+6d = 4(0.1) + 6

STEP 17

Multiply:
d=0.4+6d = 0.4 + 6

STEP 18

Add to get our final value:
d=6.4d = 6.4

SOLUTION

a. If t=1.75t = 1.75, then d=3.5d = 3.5 inches.
b. If t=4.25t = 4.25, then d=11d = 11 inches.
c. If t=2.9t = 2.9, then d=5.8d = 5.8 inches.
d. If t=3.1t = 3.1, then d=6.4d = 6.4 inches.

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