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olga2289 [7]
2 years ago
7

A gallon or unleaded gasoline weights about 6 pounds about how many ounces does 1 quart of unleaded gasoline weight hint= 1/4 ga

llon
Mathematics
2 answers:
Oliga [24]2 years ago
6 0

Answer:

24

Step-by-step explanation:

alexira [117]2 years ago
3 0

i think the answer is 1/4 because it good

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In 1 and 2, list multiples of each number to find the LCM of each pair of number.
Tanya [424]

#1

Multiples of 2:

2, 4, 6, 8, <u>10</u>, 12, 14, 16, 18, 20,...

Multiples of 5:

5, <u>10</u>, 15, 20, 25, 30, 35, 40, ...

#2

Multiples of 6:

6, 12, 18, 24, <u>30</u>, 36, 42, 48, 54, 60, ...

Multiples of 10:

10, 20, <u>30</u>, 40, 50, 60, ...

6 0
3 years ago
When Anna was born 16 years ago, her small town of Lewisville had a population of 15,000. The population has increased 11% each
Alex17521 [72]
15.000 ÷ 100

150 = 1%

1650 = 11%

1650 x 16 = 26.400

26.400 + 15.000 = 41.400

So 41.400 is the population number now.
7 0
3 years ago
Solve the following system of equations.
Yuri [45]

Answer:

x= 12

y= 10

best of luck mate

7 0
2 years ago
A pizza pan is removed at 9:00 PM from an oven whose temperature is fixed at 450°F into a room that is a constant 70°F. After 5​
gladu [14]

Answer:

A) It will get to a temperature of 125°F at 9:19 PM

B) It will get to a temperature of 150°F at 9:16 PM

C) as time passes temperature approaches the initial temperature of 450°F

Step-by-step explanation:

We are given;

Initial temperature; T_i = 450°F

Room temperature; T_r = 70°F

From Newton's law of cooling, temperature after time (t) is given as;

T(t) = T_r + (T_i - T_r)e^(-kt)

Where k is cooling rate and t is time after the initial temperature.

Now, we are told that After 5​ minutes, the temperature is 300°F.

Thus;

300 = 70 + (450 - 70)e^(-5k)

300 - 70 = 380e^(-5k)

230/380 = e^(-5k)

e^(-5k) = 0.6053

-5k = In 0.6053

-5k = -0.502

k = 0.502/5

k = 0.1004 /min

A) Thus, at temperature of 125°F, we can find the time from;

125 = 70 + (450 - 70)e^(-0.1004t)

125 - 70 = 380e^(-0.1004t)

55/380 = e^(-0.1004t)

In (55/380) = -0.1004t

-0.1004t = -1.9328

t = 1.9328/0.1018

t ≈ 19 minutes

Thus, it will get to a temperature of 125°F at 9:19 PM

B) Thus, at temperature of 150°F, we can find the time from;

150 = 70 + (450 - 70)e^(-0.1004t)

150 - 70 = 380e^(-0.1004t)

80/380 = e^(-0.1004t)

In (80/380) = -0.1004t

-0.1004t = -1.5581

t = 1.5581/0.1004

t ≈ 16 minutes.

Thus, it will get to a temperature of 150°F at 9:16 PM

C) As time passes which means as it approaches to infinity, it means that e^(-kt) gets to 1.

Thus,we have;

T(t) = T_r + (T_i - T_r)

T_r will cancel out to give;

T(t) = T_i

Thus, as time passes temperature approaches the initial temperature of 450°F

6 0
3 years ago
Can someone please help me?
docker41 [41]
THe answer is A.... 42.4 times 10.5 is 445.20
4 0
3 years ago
Read 2 more answers
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