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Math Problems
Algebra 1
Write exponential functions: word problems
The graph in the \\[yI\\]-plane shows the interest rate percentage, \\[I\\], one can expect for a savings account lasting \\[y\\] years up to a maximum of \\[\(10\)\\] years. Accounts lasting for less than \\[\(1\)\\] year are short-lived and accounts lasting for more than \\[\(5\)\\] years are long-lived. What is the significance of the \\[I\\]-intercept?
\(\newline\)A) The interest rate for a savings account lasting \(0\) years.\(\newline\)B) The interest rate for a savings account lasting \(1\) year.\(\newline\)C) The interest rate for a savings account lasting \(5\) years.\(\newline\)D) The interest rate for a savings account lasting \(10\) years.
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There are
170
170
170
deer on a reservation. The deer population is increasing at a rate of
30
%
30\%
30%
per year. Write a function that gives the deer population
P
(
t
)
P(t)
P
(
t
)
on the reservation
t
t
t
years from now.
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A biology class at Central High School predicted that a local population of animals will double in size every
12
12
12
years. The population at the beginning of
2014
2014
2014
was estimated to be
50
50
50
animals. If
P
P
P
represents the population
n
n
n
years after
2014
2014
2014
, then which of the following equations represents the class’s model of the population over time?
\newline
A)
P
=
12
+
50
n
P = 12 + 50n
P
=
12
+
50
n
\newline
B)
P
=
50
+
12
n
P = 50 + 12n
P
=
50
+
12
n
\newline
C)
P
=
50
(
2
)
12
n
P = 50(2)^{12n}
P
=
50
(
2
)
12
n
\newline
D)
P
=
50
(
2
)
n
12
P = 50(2)^{\frac{n}{12}}
P
=
50
(
2
)
12
n
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Increases in snow levels are recorded with positive numbers. Decreases in snow levels are recorded with negative numbers.
\newline
After a winter storm, the snow on Cherry Street started melting at a rate of
1
3
c
m
\frac{1}{3} \mathrm{~cm}
3
1
cm
per hour.
\newline
What was the total change in depth of the snow on Cherry Street after
3
3
3
hours?
\newline
_____cm
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Increases in temperature are recorded with positive numbers. Decreases in temperature are recorded with negative numbers.
\newline
Once the sun went down, the outside temperature decreased
2
∘
C
2^{\circ} \mathrm{C}
2
∘
C
per hour over
4
4
4
hours.
\newline
What is the total change in temperature over the
4
4
4
hours?
\newline
∘
C
{ }^{\circ} \mathrm{C}
∘
C
____
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Susan bought a new car in
2020
2020
2020
for
$
48000
\$48000
$48000
. If the value of the car decreases by
11
%
11\%
11%
each year write an exponential modle for the value of the car. Then, estimate the cost of the car in the year
2023
2023
2023
.
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This exercise uses the radioactive decay model.
\newline
The half-life of strontium
−
90
-90
−
90
is
29
29
29
years. How long (in yr) will it take a
70
70
70
-milligram sample to decay to a mass of
56
56
56
mg? (Round your answer to the nearest whole number.)
\newline
_
_
_
_
_
\_\_\_\_\_
_____
yr
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The population of a city decreases by
1
%
1 \%
1%
per year. What should we multiply the current population by to find next year's population in one step?
\newline
Answer:
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The population of a city decreases by
1
%
1 \%
1%
per year. If this year's population is
p
p
p
, which expression represents next year's population?
\newline
0.99
p
0.99 p
0.99
p
\newline
0.999
p
0.999 p
0.999
p
\newline
0.01
p
0.01 p
0.01
p
\newline
1
p
1 p
1
p
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The population of a city increases by
4
%
4 \%
4%
per year. If this year's population is
p
p
p
, which expression represents next year's population?
\newline
1.004
p
1.004 p
1.004
p
\newline
1.04
p
1.04 p
1.04
p
\newline
4
p
4 p
4
p
\newline
0.04
p
0.04 p
0.04
p
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The population of a city increases by
4
%
4 \%
4%
per year. If this year's population is
p
p
p
, which expression represents next year's population?
\newline
p
+
0.04
p
p+0.04 p
p
+
0.04
p
\newline
0.04
p
0.04 p
0.04
p
\newline
1
+
0.04
p
1+0.04 p
1
+
0.04
p
\newline
4
p
4 p
4
p
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A lake near the Arctic Circle is covered by a thick sheet of ice during the cold winter months. When spring arrives, the warm air gradually melts the ice, causing its thickness to decrease at a rate of
0.2
0.2
0.2
meters per week. After
7
7
7
weeks, the sheet is only
2.4
2.4
2.4
meters thick.
\newline
Let
y
y
y
represent the ice sheet's thickness (in meters) after
x
x
x
weeks.
\newline
Complete the equation for the relationship between the thickness and number of weeks.
\newline
y
=
□
y=\square
y
=
□
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Eagle Eye Tour Company plans to offer scenic helicopter tours as its newest attraction. The company spent
$
979
,
000
\$979,000
$979
,
000
on a pre-owned helicopter, which is expected to lose about
11
%
11\%
11%
of its value each year. Write an exponential equation in the form
y
=
a
(
b
)
x
y=a(b)^x
y
=
a
(
b
)
x
that can model the value of the helicopter,
y
y
y
,
x
x
x
years after purchase. Use whole numbers, decimals, or simplified fractions for the values of
a
a
a
and
b
b
b
.
y
=
y =
y
=
To the nearest hundred dollars, how much will the helicopter be worth
6
6
6
years after purchase?
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There are
170
170
170
deer on a reservation. The deer population is increasing at a rate of
30
%
30\%
30%
per year. Write a function that gives the deer population
p
(
t
)
p(t)
p
(
t
)
on the reservation
t
t
t
years from now.
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The next model of a sports car will cost
12.6
%
12.6 \%
12.6%
less than the current model. The current model costs
$
58
,
000
\$ 58,000
$58
,
000
. How much will the price decrea will be the price of the next model?
\newline
Decrease in price:
$
□
\quad \$ \square
$
□
\newline
Price of next model:
$
□
\quad \$ \square
$
□
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A radioactive compound with mass
330
330
330
grams decays at a rate of
21
%
21 \%
21%
per hour. Which equation represents how many grams of the compound will remain after
4
4
4
hours?
\newline
C
=
330
(
1
−
0.21
)
C=330(1-0.21)
C
=
330
(
1
−
0.21
)
\newline
C
=
330
(
1.21
)
4
C=330(1.21)^{4}
C
=
330
(
1.21
)
4
\newline
C
=
330
(
1
+
0.21
)
4
C=330(1+0.21)^{4}
C
=
330
(
1
+
0.21
)
4
\newline
C
=
330
(
1
−
0.21
)
(
1
−
0.21
)
(
1
−
0.21
)
(
1
−
0.21
)
C=330(1-0.21)(1-0.21)(1-0.21)(1-0.21)
C
=
330
(
1
−
0.21
)
(
1
−
0.21
)
(
1
−
0.21
)
(
1
−
0.21
)
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Chloe has a collection of vintage action figures that is worth
$
310
\$ 310
$310
. If the collection appreciates at a rate of
15
%
15 \%
15%
per year, which equation represents the value of the collection after
5
5
5
years?
\newline
V
=
310
(
1.15
)
5
V=310(1.15)^{5}
V
=
310
(
1.15
)
5
\newline
V
=
310
(
1
+
0.15
)
V=310(1+0.15)
V
=
310
(
1
+
0.15
)
\newline
V
=
310
(
1
−
0.15
)
5
V=310(1-0.15)^{5}
V
=
310
(
1
−
0.15
)
5
\newline
V
=
310
(
0.85
)
5
V=310(0.85)^{5}
V
=
310
(
0.85
)
5
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A radioactive compound with mass
280
280
280
grams decays at a rate of
18.3
%
18.3 \%
18.3%
per hour. Which equation represents how many grams of the compound will remain after
2
2
2
hours?
\newline
C
=
280
(
1.183
)
2
C=280(1.183)^{2}
C
=
280
(
1.183
)
2
\newline
C
=
280
(
1
−
0.183
)
(
1
−
0.183
)
C=280(1-0.183)(1-0.183)
C
=
280
(
1
−
0.183
)
(
1
−
0.183
)
\newline
C
=
280
(
1
+
0.183
)
2
C=280(1+0.183)^{2}
C
=
280
(
1
+
0.183
)
2
\newline
C
=
280
(
1
−
0.183
)
(
1
−
0.183
)
(
1
−
0.183
)
(
1
−
0.183
)
C=280(1-0.183)(1-0.183)(1-0.183)(1-0.183)
C
=
280
(
1
−
0.183
)
(
1
−
0.183
)
(
1
−
0.183
)
(
1
−
0.183
)
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In a lab experiment, a population of
250
250
250
bacteria is able to triple every hour. Which equation matches the number of bacteria in the population after
2
2
2
hours?
\newline
B
=
250
(
3
)
2
B=250(3)^{2}
B
=
250
(
3
)
2
\newline
B
=
3
(
250
)
2
B=3(250)^{2}
B
=
3
(
250
)
2
\newline
B
=
3
(
250
)
(
250
)
B=3(250)(250)
B
=
3
(
250
)
(
250
)
\newline
B
=
250
(
3
)
(
3
)
(
3
)
(
3
)
B=250(3)(3)(3)(3)
B
=
250
(
3
)
(
3
)
(
3
)
(
3
)
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A group consisting of
8
8
8
aggressive zombies triples in size every hour. Which equation matches the number of zombies after
5
5
5
hours?
\newline
Z
=
3
(
1
+
8
)
5
Z=3(1+8)^{5}
Z
=
3
(
1
+
8
)
5
\newline
Z
=
3
(
8
)
5
Z=3(8)^{5}
Z
=
3
(
8
)
5
\newline
Z
=
8
(
3
)
5
Z=8(3)^{5}
Z
=
8
(
3
)
5
\newline
Z
=
8
(
3
)
Z=8(3)
Z
=
8
(
3
)
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A radioactive compound with mass
420
420
420
grams decays at a rate of
4
%
4 \%
4%
per hour. Which equation represents how many grams of the compound will remain after
5
5
5
hours?
\newline
C
=
420
(
0.6
)
5
C=420(0.6)^{5}
C
=
420
(
0.6
)
5
\newline
C
=
420
(
1
+
0.04
)
5
C=420(1+0.04)^{5}
C
=
420
(
1
+
0.04
)
5
\newline
C
=
420
(
1
−
0.04
)
(
1
−
0.04
)
(
1
−
0.04
)
C=420(1-0.04)(1-0.04)(1-0.04)
C
=
420
(
1
−
0.04
)
(
1
−
0.04
)
(
1
−
0.04
)
\newline
C
=
420
(
0.96
)
5
C=420(0.96)^{5}
C
=
420
(
0.96
)
5
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A town has a population of
36
36
36
,
500
500
500
and grows at a rate of
6.4
%
6.4 \%
6.4%
every year. Which equation represents the town's population after
5
5
5
years?
\newline
P
=
36
,
500
(
1
+
0.064
)
P=36,500(1+0.064)
P
=
36
,
500
(
1
+
0.064
)
\newline
P
=
36
,
500
(
1
−
0.064
)
5
P=36,500(1-0.064)^{5}
P
=
36
,
500
(
1
−
0.064
)
5
\newline
P
=
36
,
500
(
0.936
)
5
P=36,500(0.936)^{5}
P
=
36
,
500
(
0.936
)
5
\newline
P
=
36
,
500
(
1
+
0.064
)
5
P=36,500(1+0.064)^{5}
P
=
36
,
500
(
1
+
0.064
)
5
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Kamron's house is
0.5
0.5
0.5
mile from the grocery store. If Kamron's house is
1.5
1.5
1.5
miles closer to the grocery store than Amy's house is, which of the following can be used to determine
x
x
x
, the distance between Amy's house and the grocery store in miles?
\newline
Choose
1
1
1
answer:
\newline
(A)
0.5
×
x
=
1.5
0.5 \times x=1.5
0.5
×
x
=
1.5
\newline
(B)
1.5
−
x
=
0.5
1.5-x=0.5
1.5
−
x
=
0.5
\newline
(C)
x
−
1.5
=
0.5
x-1.5=0.5
x
−
1.5
=
0.5
\newline
(D)
x
+
0.5
=
1.5
x+0.5=1.5
x
+
0.5
=
1.5
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Lakewood Children's Theater started a summer program last year with
245
245
245
students. Thanks to some advertising during the off-season,
294
294
294
students have enrolled this year. The theater will continue this advertising strategy in hopes that enrollment will continue to increase.
\newline
Write an exponential equation in the form
y
=
a
(
b
)
x
y=a(b)^{x}
y
=
a
(
b
)
x
that can model the number of enrolled students,
y
,
x
y, x
y
,
x
years after the program began.
\newline
Use whole numbers, decimals, or simplified fractions for the values of
a
a
a
and
b
b
b
.
\newline
y
=
y=
y
=
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In
1950
1950
1950
, the per capita gross domestic product (GDP) of Australia was approximately
$
1800
\$1800
$1800
. Each year afterwards, the per capita GDP increased by approximately
6.7
%
6.7\%
6.7%
. Write a function that gives the approximate per capita GDP
G
(
t
)
G(t)
G
(
t
)
of Australia
t
t
t
years after.
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Kamron's house is
0.5
0.5
0.5
mile from the grocery store. If Kamron's house is
1.5
1.5
1.5
miles closer to the grocery store than Amy's house is, which of the following can be used to determine
x
x
x
, the distance between Amy's house and the grocery store in miles?
\newline
Choose
1
1
1
answer:
\newline
(A)
0.5
×
x
=
1.5
0.5 \times x=1.5
0.5
×
x
=
1.5
\newline
(B)
1.5
−
x
=
0.5
1.5-x=0.5
1.5
−
x
=
0.5
\newline
(C)
x
−
1.5
=
0.5
x-1.5=0.5
x
−
1.5
=
0.5
\newline
(D)
x
+
0.5
=
1.5
x+0.5=1.5
x
+
0.5
=
1.5
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The rate of change
d
P
d
t
\frac{d P}{d t}
d
t
d
P
of the number of wolves at a national park is modeled by a logistic differential equation. The maximum capacity of the park is
955
955
955
wolves. At
8
8
8
PM, the number of wolves at the national park is
218
218
218
and is increasing at a rate of
26
26
26
wolves per day. Write a differential equation to describe the situation.
\newline
d
P
d
t
=
□
\frac{d P}{d t}=\square
d
t
d
P
=
□
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The rate of change
d
P
d
t
\frac{d P}{d t}
d
t
d
P
of the number of people infected by a disease is modeled by a logistic differential equation. The maximum capacity of the village is
943
943
943
people. At
8
8
8
PM, the number of people infected by the disease is
106
106
106
and is increasing at a rate of
30
30
30
people per hour. Write a differential equation to describe the situation.
\newline
d
P
d
t
=
□
\frac{d P}{d t}=\square
d
t
d
P
=
□
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The rate of change
d
P
d
t
\frac{d P}{d t}
d
t
d
P
of the number of deer on an island is modeled by a logistic differential equation. The maximum capacity of the island is
541
541
541
deer. At
12
P
M
12 \mathrm{PM}
12
PM
, the number of deer on the island is
228
228
228
and is increasing at a rate of
15
15
15
deer per day. Write a differential equation to describe the situation.
\newline
d
P
d
t
=
□
\frac{d P}{d t}=\square
d
t
d
P
=
□
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The rate of change
d
P
d
t
\frac{d P}{d t}
d
t
d
P
of the number of deer on an island is modeled by a logistic differential equation. The maximum capacity of the island is
673
673
673
deer. At
8
P
M
8 \mathrm{PM}
8
PM
, the number of deer on the island is
172
172
172
and is increasing at a rate of
37
37
37
deer per day. Write a differential equation to describe the situation.
\newline
d
P
d
t
=
□
\frac{d P}{d t}=\square
d
t
d
P
=
□
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The rate of change
d
P
d
t
\frac{d P}{d t}
d
t
d
P
of the number of fox at a national park is modeled by a logistic differential equation. The maximum capacity of the park is
896
896
896
fox. At
8
8
8
PM, the number of fox at the national park is
184
184
184
and is increasing at a rate of
28
28
28
fox per day. Write a differential equation to describe the situation.
\newline
d
P
d
t
=
□
\frac{d P}{d t}=\square
d
t
d
P
=
□
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The rate of change
d
P
d
t
\frac{d P}{d t}
d
t
d
P
of the number of students who heard a rumor is modeled by a logistic differential equation. The maximum capacity of the school is
942
942
942
students. At
5
A
M
5 \mathrm{AM}
5
AM
, the number of students who heard the rumor is
233
233
233
and is increasing at a rate of
37
37
37
students per hour. Write a differential equation to describe the situation.
\newline
d
P
d
t
=
□
\frac{d P}{d t}=\square
d
t
d
P
=
□
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The rate of change
d
P
d
t
\frac{d P}{d t}
d
t
d
P
of the number of deer on an island is modeled by a logistic differential equation. The maximum capacity of the island is
593
593
593
deer. At
5
P
M
5 \mathrm{PM}
5
PM
, the number of deer on the island is
210
210
210
and is increasing at a rate of
16
16
16
deer per day. Write a differential equation to describe the situation.
\newline
d
P
d
t
=
□
\frac{d P}{d t}=\square
d
t
d
P
=
□
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The rate of change
d
P
d
t
\frac{d P}{d t}
d
t
d
P
of the number of people on an island is modeled by a logistic differential equation. The maximum capacity of the island is
745
745
745
people. At
11
11
11
AM, the number of people on the island is
186
186
186
and is increasing at a rate of
24
24
24
people per hour. Write a differential equation to describe the situation.
\newline
d
P
d
t
=
□
\frac{d P}{d t}=\square
d
t
d
P
=
□
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Wangari plants trees at a constant rate of
12
12
12
trees every
3
3
3
hours.
\newline
Write an equation that relates
p
p
p
, the number of trees Wangari plants, and
h
h
h
, the time she spends planting them in hours.
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Mia tried to solve an equation step by step.
\newline
0.1
t
−
3
=
5
0.1 t-3=5
0.1
t
−
3
=
5
\newline
0.1
t
=
2
0.1 t=2 \quad
0.1
t
=
2
Step
1
1
1
\newline
t
=
20
Step
2
t=20 \quad \text { Step } 2
t
=
20
Step
2
\newline
Find Mia's mistake.
\newline
Choose
1
1
1
answer:
\newline
(A) Step
1
1
1
\newline
(B) Step
2
2
2
\newline
(C) Mia did not make a mistake.
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Kwame must earn more than
16
16
16
stars per day to get a prize from the classroom treasure box.
\newline
Write an inequality that describes
S
S
S
, the number of stars Kwame must earn per day to get a prize from the classroom treasure box.
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Naira paddled
4.5
k
m
4.5 \mathrm{~km}
4.5
km
with the current in the same amount of time as it took her to paddle
3
k
m
3 \mathrm{~km}
3
km
against the current. Naira paddled at an average rate of
5
k
m
h
5 \frac{\mathrm{km}}{\mathrm{h}}
5
h
km
relative to the water each way. Assume the speed of the current was constant.
\newline
What was the speed of the current?
\newline
□
k
m
h
\square \frac{\mathrm{km}}{\mathrm{h}}
□
h
km
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After a special medicine is introduced into a petri dish containing a bacterial culture, the number of bacteria remaining in the dish decreases rapidly.
\newline
The population loses
1
4
\frac{1}{4}
4
1
of its size every
44
44
44
seconds. The number of remaining bacteria can be modeled by a function,
N
N
N
, which depends on the amount of time,
t
t
t
(in seconds).
\newline
Before the medicine was introduced, there were
11
11
11
,
880
880
880
bacteria in the Petri dish.
\newline
Write a function that models the number of remaining bacteria
t
t
t
seconds since the medicine was introduced.
\newline
N
(
t
)
=
□
N(t)=\square
N
(
t
)
=
□
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Nicholas sent a chain letter to his friends, asking them to forward the letter to more friends. Every
12
12
12
weeks, the number of people who receive the email increases by an additional
99
%
99 \%
99%
, and can be modeled by a function,
P
P
P
, which depends on the amount of time,
t
t
t
(in weeks).
\newline
Nicholas initially sent the chain letter to
50
50
50
friends.
\newline
Write a function that models the number of people who receive the email
t
t
t
weeks since Nicholas initially sent the chain letter.
\newline
P
(
t
)
=
□
P(t)=\square
P
(
t
)
=
□
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On the first day of spring, an entire field of flowering trees blossoms. The population of locusts consuming these flowers rapidly increases as the trees blossom. The locust population gains
87
%
87 \%
87%
of its size every
2
2
2
.
4
4
4
days, and can be modeled by a function,
L
L
L
, which depends on the amount of time,
t
t
t
(in days).
\newline
Before the first day of spring, there were
1100
1100
1100
locusts in the population.
\newline
Write a function that models the locust population
t
t
t
days since the first day of spring.
\newline
L
(
t
)
=
L(t)=
L
(
t
)
=
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Carbon
−
14
-14
−
14
is an element which loses exactly half of its mass every
5730
5730
5730
years. The mass of a sample of carbon-
14
14
14
can be modeled by a function,
M
M
M
, which depends on its age,
t
t
t
(in years).
\newline
We measure that the initial mass of a sample of carbon
−
14
-14
−
14
is
741
741
741
grams.
\newline
Write a function that models the mass of the carbon-
14
14
14
sample remaining
t
t
t
years since the initial measurement.
\newline
M
(
t
)
=
□
M(t)=\square
M
(
t
)
=
□
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Lenmana started studying how the number of branches on her tree grows over time. The number of branches increases by a factor of
11
6
\frac{11}{6}
6
11
every
3
3
3
.
5
5
5
years, and can be modeled by a function,
N
N
N
, which depends on the amount of time,
t
t
t
(in years).
\newline
When Lenmana began the study, her tree had
48
48
48
branches.
\newline
Write a function that models the number of branches
t
t
t
years since Lenmana began studying her tree.
\newline
N
(
t
)
=
□
N(t)=\square
N
(
t
)
=
□
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Bela started studying how the number of branches on her tree grows over time. Every
2
2
2
.
9
9
9
years, the number of branches increases by an additional
83
%
83 \%
83%
, and can be modeled by a function,
N
N
N
, which depends on the amount of time,
t
t
t
(in years).
\newline
When Bela began the study, her tree had
60
60
60
branches.
\newline
Write a function that models the number of branches
t
t
t
years since Bela began studying her tree.
\newline
N
(
t
)
=
□
N(t)=\square
N
(
t
)
=
□
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Carbon
−
14
-14
−
14
is an element which loses
1
10
\frac{1}{10}
10
1
of its mass every
871
871
871
years. The mass of a sample of carbon
−
14
-14
−
14
can be modeled by a function,
M
M
M
, which depends on its age,
t
t
t
(in years).
\newline
We measure that the initial mass of a sample of carbon
−
14
-14
−
14
is
960
960
960
grams.
\newline
Write a function that models the mass of the carbon-
14
14
14
sample remaining
t
t
t
years since the initial measurement.
\newline
M
(
t
)
=
M(t)=
M
(
t
)
=
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Aleksandra started studying how the number of branches on her tree grows over time. Every
1
1
1
.
5
5
5
years, the number of branches increases by an addition of
2
7
\frac{2}{7}
7
2
of the total number of branches. The number of branches can be modeled by a function,
N
N
N
, which depends on the amount of time,
t
t
t
(in years).
\newline
When Aleksandra began the study, her tree had
52
52
52
branches.
\newline
Write a function that models the number of branches
t
t
t
years since Aleksandra began studying her tree.
\newline
N
(
t
)
=
□
N(t)=\square
N
(
t
)
=
□
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The moon's illumination changes in a periodic way that can be modeled by a trigonometric function.
\newline
On the night of a full moon, the moon provides about
0
0
0
.
25
25
25
lux of illumination (lux is the SI unit of illuminance). During a new moon, the moon provides
0
0
0
lux of illumination. The period of the lunar cycle is
29
29
29
.
53
53
53
days long. The moon will be full on December
25
25
25
,
2015
2015
2015
. Note that December
25
25
25
is
7
7
7
days before January
1
1
1
.
\newline
Find the formula of the trigonometric function that models the illumination
L
L
L
of the moon
t
t
t
days after January
1
1
1
,
2016
2016
2016
. Define the function using radians.
\newline
L
(
t
)
=
□
L(t)=\square
L
(
t
)
=
□
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Valeria is playing with her accordion.
\newline
The length of the accordion
A
(
t
)
A(t)
A
(
t
)
(in
c
m
\mathrm{cm}
cm
) after she starts playing as a function of time
t
t
t
(in seconds) can be modeled by a sinusoidal expression of the form
a
⋅
cos
(
b
⋅
t
)
+
d
a \cdot \cos (b \cdot t)+d
a
⋅
cos
(
b
⋅
t
)
+
d
.
\newline
At
t
=
0
t=0
t
=
0
, when she starts playing, the accordion is
15
c
m
15 \mathrm{~cm}
15
cm
long, which is the shortest it gets.
1
1
1
.
5
5
5
seconds later the accordion is at its average length of
21
c
m
21 \mathrm{~cm}
21
cm
.
\newline
Find
A
(
t
)
A(t)
A
(
t
)
.
\newline
t
t
t
should be in radians.
\newline
A
(
t
)
=
□
A(t)=\square
A
(
t
)
=
□
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Kaliska is jumping rope.
\newline
The vertical height of the center of her rope off the ground
R
(
t
)
R(t)
R
(
t
)
(in
c
m
\mathrm{cm}
cm
) as a function of time
t
t
t
(in seconds) can be modeled by a sinusoidal expression of the form
a
⋅
cos
(
b
⋅
t
)
+
d
a \cdot \cos (b \cdot t)+d
a
⋅
cos
(
b
⋅
t
)
+
d
.
\newline
At
t
=
0
t=0
t
=
0
, when she starts jumping, her rope is
0
c
m
0 \mathrm{~cm}
0
cm
off the ground, which is the minimum. After
π
12
\frac{\pi}{12}
12
π
seconds, it reaches a height of
60
c
m
60 \mathrm{~cm}
60
cm
from the ground, which is half of its maximum height.
\newline
Find
R
(
t
)
R(t)
R
(
t
)
.
\newline
t
t
t
should be in radians.
\newline
R
(
t
)
=
□
R(t)=\square
R
(
t
)
=
□
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The day length in Manila, Philippines, varies over time in a periodic way that can be modeled by a trigonometric function.
\newline
Assume the length of the year (which is the period of change) is exactly
365
365
365
days long. The shortest day of the year is December
21
21
21
, and it's
675
675
675
.
85
85
85
minutes long. Manila's longest day is
779
779
779
.
60
60
60
minutes long. Note that December
21
21
21
is
11
11
11
days before January
1
1
1
.
\newline
Find the formula of the trigonometric function that models the length
L
L
L
of the day
t
t
t
days after January
1
1
1
. Define the function using radians.
\newline
L
(
t
)
=
□
L(t)=\square
L
(
t
)
=
□
\newline
What is the day length on the People Power Anniversary (February
25
25
25
, which is
55
55
55
days after January
1
1
1
) in Manila? Round your answer, if necessary, to two decimal places.
\newline
minutes
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