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Math Problems
Grade 8
Convert between Celsius and Fahrenheit
Convert the temperature from degrees Celsius to degrees Fahrenheit, using the formula below.
\newline
F
=
(
9
5
×
C
)
+
32
F = \left(\frac{9}{5} \times C\right) + 32
F
=
(
5
9
×
C
)
+
32
\newline
100
°
C
=
_
_
_
_
_
°
F
100\degree C = \_\_\_\_\_\degree F
100°
C
=
_____°
F
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Convert the temperature from degrees Celsius to degrees Fahrenheit, using the formula below.
\newline
F
=
(
9
5
×
C
)
+
32
F = \left(\frac{9}{5} \times C\right) + 32
F
=
(
5
9
×
C
)
+
32
\newline
3
0
∘
C
=
30^\circ\text{C} =
3
0
∘
C
=
_____
∘
F
^\circ\text{F}
∘
F
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Convert the temperature from degrees Fahrenheit to degrees Celsius, using the formula below.
\newline
C
=
5
9
×
(
F
−
32
)
C = \frac{5}{9} \times (F - 32)
C
=
9
5
×
(
F
−
32
)
\newline
11
3
∘
F
=
_
_
_
_
_
_
_
_
_
∘
C
113^\circ F = \_\_\_\_\_\_\_\_\_\circ C
11
3
∘
F
=
_________
∘
C
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Convert the temperature from degrees Fahrenheit to degrees Celsius, using the formula below.
\newline
C
=
5
9
×
(
F
−
32
)
C = \frac{5}{9} \times (F - 32)
C
=
9
5
×
(
F
−
32
)
\newline
167
°
F
=
_
_
_
_
_
°
C
167\degree F = \_\_\_\_\_\degree C
167°
F
=
_____°
C
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Convert the temperature from degrees Fahrenheit to degrees Celsius, using the formula below.
\newline
C
=
5
9
×
(
F
−
32
)
C = \frac{5}{9} \times (F - 32)
C
=
9
5
×
(
F
−
32
)
\newline
41
°
F
=
_
_
_
_
_
°
C
41\degree F = \_\_\_\_\_\degree C
41°
F
=
_____°
C
Get tutor help
Convert the temperature from degrees Fahrenheit to degrees Celsius, using the formula below.
\newline
C
=
5
9
×
(
F
−
32
)
C = \frac{5}{9} \times (F - 32)
C
=
9
5
×
(
F
−
32
)
\newline
86
°
F
=
_
_
_
_
_
°
C
86\degree F = \_\_\_\_\_\degree C
86°
F
=
_____°
C
Get tutor help
Convert the temperature from degrees Celsius to degrees Fahrenheit, using the formula below.
\newline
F
=
(
9
5
×
C
)
+
32
F = \left(\frac{9}{5} \times C\right) + 32
F
=
(
5
9
×
C
)
+
32
\newline
95
°
C
=
_
_
_
_
_
°
F
95\degree C = \_\_\_\_\_\degree F
95°
C
=
_____°
F
Get tutor help
Convert the temperature from degrees Fahrenheit to degrees Celsius, using the formula below.
\newline
C
=
5
9
×
(
F
−
32
)
C = \frac{5}{9} \times (F - 32)
C
=
9
5
×
(
F
−
32
)
\newline
203
°
F
=
_
_
_
_
_
°
C
203\degree F = \_\_\_\_\_\degree C
203°
F
=
_____°
C
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Convert the temperature from degrees Celsius to degrees Fahrenheit, using the formula below.
\newline
F
=
(
9
5
×
C
)
+
32
F = \left(\frac{9}{5} \times C\right) + 32
F
=
(
5
9
×
C
)
+
32
\newline
5
°
C
=
5\degree C =
5°
C
=
_____
°
F
\degree F
°
F
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Convert the temperature from degrees Fahrenheit to degrees Celsius, using the formula below.
\newline
C
=
5
9
×
(
F
−
32
)
C = \frac{5}{9} \times (F - 32)
C
=
9
5
×
(
F
−
32
)
\newline
59
°
F
=
_
_
_
_
°
C
59\degree F = \_\_\_\_\degree C
59°
F
=
____°
C
Get tutor help
Convert the temperature from degrees Celsius to degrees Fahrenheit, using the formula below.
\newline
F
=
(
9
5
×
C
)
+
32
F = \left(\frac{9}{5} \times C\right) + 32
F
=
(
5
9
×
C
)
+
32
\newline
6
5
∘
C
=
65^\circ\text{C} =
6
5
∘
C
=
_____
∘
F
^\circ\text{F}
∘
F
Get tutor help
Convert the temperature from degrees Celsius to degrees Fahrenheit, using the formula below.
\newline
F
=
(
9
5
×
C
)
+
32
F = \left(\frac{9}{5} \times C\right) + 32
F
=
(
5
9
×
C
)
+
32
\newline
75
°
C
=
_
_
_
_
_
°
F
75\degree C = \_\_\_\_\_\degree F
75°
C
=
_____°
F
Get tutor help
Convert the temperature from degrees Celsius to degrees Fahrenheit, using the formula below.
\newline
F
=
(
9
5
×
C
)
+
32
F = \left(\frac{9}{5} \times C\right) + 32
F
=
(
5
9
×
C
)
+
32
\newline
10
°
C
=
10\degree C =
10°
C
=
_____
°
F
\degree F
°
F
Get tutor help
Convert the temperature from degrees Fahrenheit to degrees Celsius, using the formula below.
\newline
C
=
5
9
×
(
F
−
32
)
C = \frac{5}{9} \times (F - 32)
C
=
9
5
×
(
F
−
32
)
\newline
50
°
F
=
_
_
_
_
_
°
C
50\degree F = \_\_\_\_\_\degree C
50°
F
=
_____°
C
Get tutor help
Convert the temperature from degrees Celsius to degrees Fahrenheit, using the formula below.
\newline
F
=
(
9
5
×
C
)
+
32
F = \left(\frac{9}{5} \times C\right) + 32
F
=
(
5
9
×
C
)
+
32
\newline
50
°
C
=
_
_
_
_
_
°
F
50\degree C = \_\_\_\_\_\degree F
50°
C
=
_____°
F
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The expression
100
+
20
m
100+20m
100
+
20
m
gives the volume of water in Eduardo's pool (in liters) after Eduardo spends
\newline
m
m
m
minutes filling his pool.
\newline
What is the volume of water in Eduardo's pool after he fills it for
5
1
4
5\frac{1}{4}
5
4
1
minutes?
\newline
□
\square
□
liters
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A study by the department of education of a certain state was trying to determine the mean SAT scores of the graduating high school seniors. The study found that the mean SAT score was
457
457
457
with a margin of error of
20
20
20
. Which of the following is a reasonable value for the true mean SAT score of graduating high school seniors?
\newline
471
471
471
.
2
2
2
\newline
485
485
485
.
2
2
2
\newline
428
428
428
\newline
432
432
432
.
8
8
8
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Temperature can be measured in two different common units; degrees Celsius and degrees Fahrenheit.
f
f
f
represents the temperature in degrees Fahrenheit as a function of the temperature
c
c
c
in degrees Celsius.
\newline
f
=
32
+
1.8
c
f = 32 + 1.8c
f
=
32
+
1.8
c
\newline
What is the temperature increase in degrees Fahrenheit that is equivalent to a temperature increase by
10
10
10
degrees Celsius?
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5
9
(
F
−
32
)
\frac{5}{9}(F - 32)
9
5
(
F
−
32
)
The equation above shows how temperature
F
F
F
, measured in degrees Fahrenheit, relates to a temperature
C
C
C
, measured in degrees Celsius. Based on the equation, which of the following must be true?
\newline
A temperature increase of
1
1
1
degree Fahrenheit is equivalent to a temperature increase of
5
9
\frac{5}{9}
9
5
degree Celsius.
\newline
A temperature increase of
1
1
1
degree Celsius is equivalent to a temperature increase of
1.8
1.8
1.8
degrees Fahrenheit.
\newline
A temperature increase of
5
9
\frac{5}{9}
9
5
degree Fahrenheit is equivalent to a temperature increase of
1
1
1
degree Celsius.
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The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at
\newline
300
K
300\,\text{K}
300
K
is :
\newline
(Atomic masses:
\newline
He
=
4
u
\text{He}=4\,\text{u}
He
=
4
u
,
O
=
16
u
\text{O}=16\,\text{u}
O
=
16
u
)
\newline
(
1
1
1
)
\newline
1200
K
1200\,\text{K}
1200
K
\newline
(
2
2
2
)
\newline
600
K
600\,\text{K}
600
K
\newline
(
3
3
3
)
\newline
300
K
300\,\text{K}
300
K
\newline
(
4
4
4
)
\newline
2400
K
2400\,\text{K}
2400
K
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On a
40
Fahrenheit
40 \, \text{Fahrenheit}
40
Fahrenheit
day, a
25
mile per hour
25 \, \text{mile per hour}
25
mile per hour
wind creates a wind chill of
29
Fahrenheit
29 \, \text{Fahrenheit}
29
Fahrenheit
. To approximate the wind chill at various temperatures caused by a
25
mile per hour
25 \, \text{mile per hour}
25
mile per hour
wind, Elon uses this rule: take
24
24
24
less than
4
3
\frac{4}{3}
3
4
of the air temperature(
Fahrenheit
\text{Fahrenheit}
Fahrenheit
). If a
25
mile per hour
25 \, \text{mile per hour}
25
mile per hour
wind creates a chill of
−
5
Fahrenheit
-5 \, \text{Fahrenheit}
−
5
Fahrenheit
, which of the following best approximates the corresponding air temperature?
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The angle below measures
5
5
5
.
9
9
9
radians, and the circle centered at the angle's vertex has a radius
2
2
2
.
2
2
2
units long.
\newline
Determine the exact coordinates of the terminal point
(
x
,
y
)
(x, y)
(
x
,
y
)
.
\newline
-
x
=
x=
x
=
\newline
Preview
\newline
-
y
=
y=
y
=
\newline
Preview
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In January, the average temperature
t
t
t
hours after midnight in Mumbai, India, is given by
\newline
T
(
t
)
=
24.5
−
5.5
sin
(
2
π
(
t
+
1
)
24
)
.
T(t)=24.5-5.5 \sin \left(\frac{2 \pi(t+1)}{24}\right) .
T
(
t
)
=
24.5
−
5.5
sin
(
24
2
π
(
t
+
1
)
)
.
\newline
What is the coldest time of day in Mumbai? Give an exact answer.
\newline
□
\square
□
hours after midnight
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Temperature can be measured in two different common units; degrees Celsius and degrees Fahrenheit.
f
f
f
represents the temperature in degrees Fahrenheit as a function of the temperature
c
c
c
in degrees Celsius.
f
=
32
+
1.8
c
f = 32 + 1.8c
f
=
32
+
1.8
c
What is the temperature increase in degrees Fahrenheit that is equivalent to a temperature increase by
10
10
10
degrees Celsius?
Get tutor help
