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Naddik [55]
3 years ago
10

Represent each of these temperatures in degrees Fahrenheit with a positive or negative number.

Mathematics
1 answer:
Anuta_ua [19.1K]3 years ago
5 0
5 degrees above is 5°F
3 degrees below is -3°F
6 degrees above is 6°F
2 3/4 below is -2.74°F
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I need he answer pls answer fast answer don’t scam
Viktor [21]

Answer:

30 ft²

Step-by-step explanation:

Area of trapezium = 1/2 (a + b)h

a = 6 ft , b = 9 ft , h = 4 ft

Area = 1/2 (6 + 9) 4

= 30 ft²

6 0
2 years ago
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Keenan will run 2.5 miles from his house to Jared’s house. He plans to hang out for 45 minutes before walking home. If he can ru
jenyasd209 [6]

Answer:

Kennan will be from home approximately an hour and 48 minutes.

Step-by-step explanation:

We must know that total time (t_{T}) that Keenan will be from home is the sum of run (t_{R}), hang out (t_{H}) and walk times (t_{W}), measured in hours:

t_{T} = t_{R}+t_{H}+t_{W}

If Keenan runs and walks at constant speed, then equation above can be expanded:

t_{T} = \frac{x_{R}}{v_{R}}+t_{H}+ \frac{x_{W}}{v_{W}}

Where:

x_{R}, x_{W} - Run and walk distances, measured in miles.

v_{R}, v_{W} - Run and walk speeds, measured in miles per hour.

Given that x_{R}=x_{W} = 2.5\,mi, v_{R} = 6\,\frac{mi}{h}, v_{W} = 4\,\frac{mi}{h} and t_{H} = 0.75\,h, the total time is:

t_{T} = \frac{2.5\,mi}{6\,\frac{mi}{h} } + 0.75\,h+\frac{2.5\,mi}{4\,\frac{mi}{h} }

t_{T} = 1.792\,h (1\,h\,48\,m)

Kennan will be from home approximately an hour and 48 minutes.

7 0
3 years ago
The Aquarium charges $16.00 for admission and $2.50 for each ride ticket. Mountain Rides charges $25.00 for admission and $0.25
Darina [25.2K]

Answer:

B

Step-by-step explanation:

Because 16 + 2.50 is less than 25 + 0.25x

7 0
2 years ago
How many 1/3 cups are rings are there in 12 4/6 cups of juice​
Bingel [31]

Answer:

how many 1/3 cups are rings are there in 12 4/6 cups of juice​?

12 of 4/6 cup of juice= 12 x 4/6= 8

1/3 cups of rings is there in 8 cups of juice, then we have

1/3 x 8= 8/3

Step-by-step explanation:

5 0
3 years ago
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A fishing tackle box is13 inches long, 6 inches wide, and 2.5 inches high. What is the volume of the tackle box?
V125BC [204]
13 times 6 times 2.5 equals 195 square inches
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