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lawyer [7]
3 years ago
7

you run 10 miles in 1 hour and 15 mintues keeping the same pance how long would it take to run 15miles

Mathematics
2 answers:
lakkis [162]3 years ago
7 0

Answer:

it would take you 1 hour and 30 mins

Step-by-step explanation:

it would be that because 1 hour is 60 mins and it takes you 60 mins to run 1 mile add half a mil and then it would be 1 hr 30

Montano1993 [528]3 years ago
4 0

Answer:

1 hour and 30 minutes

Step-by-step explanation:

because it took an hour for 10 mile so add another five miles and it would take 30 minutes

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P+7 – 5 = –10 – 3p<br><br> Step by step
oksian1 [2.3K]

Step-by-step explanation:

you add the three p to p and that's 4p and

4p+7-5=-10

4p-2=-10

+2=+2

4p=-8

devide by four on both side so p is-2

6 0
3 years ago
Cameron has 2 video game holder stands each can hold 2 rows of 20 games write and evaluate a numerical expression to find the to
Aleksandr [31]

Answer:

so this would be the equation

2(2*10)=v

So the number on the outside is the number of stands, the second 2 is for the two rows, and the 10 is the games. I did this because its two rows and combined they make 20 games. It took me a little to figure that out.

This shows the number of games total in both holders.


5 0
3 years ago
$43.68532 rounded to the nearest dollar is?
Sedaia [141]

Note that when rounding, if the number is:

5 or greater, round up

4 or less, round down

43.68532 rounded to the nearest dollar is 44, as 6 rounds up

$44 is your answer

hope this helps

8 0
3 years ago
1. How much heat is absorbed by a 50g iron skillet when its temperature rises from 10oC to 124oC? Joules
hoa [83]
<span>1. How much heat is absorbed by a 50g iron skillet when its temperature rises from 10oC to 124oC? Joules

Formula: Heat = mass * specific heat * ΔT

Data:
mass = 50 g = 0.050 kg
specific heat of iron = 450 J/ kg °C
ΔT = 124°C - 10°C ¿ 114 °C

=> heat = 0.050kg * 450 J / kg°C * 114°C ≈ 2.6 J

2. If a refrigerator is a heat pump that follows the first law of thermodynamics, how much heat was removed from food inside of the refrigerator if it released 492J of energy to the room? Joules

The firs law of thermodynamics is conservation of energy => energy removed from inside of the refrigerator = energy released to the room

=> Answer = 492 J

3. How much heat is needed to raise the temperature of 45g of water by 63oC? Joules

Formula: heat = mass * specific heat * ΔT

specific heat of water = 4186 J / Kg °C

heat = 0.045 kg * 4186 J/kg °C * 63°C = 11,867.31 J
</span>
7 0
3 years ago
Solve for x please help
Fed [463]

Answer:

x = 5

Step-by-step explanation:

By remote exterior angle theorem:

(6 + 25x) \degree = (18x - 3)\degree  + 44\degree  \\ 6 + 25x = 18x - 3 + 44 \\ 25x - 18x = 41 - 6 \\ 7x = 35 \\ x =  \frac{35}{7}  \\ \huge \red{ \boxed{ x = 5}}

3 0
3 years ago
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