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mina [271]
3 years ago
13

3/5 of your job is completed. What fraction of the job remains to be done?

Mathematics
2 answers:
Sphinxa [80]3 years ago
7 0
2/5. 2/5 + 3/5 = 5/5
Sergio [31]3 years ago
7 0

Hi!

<h3>You need to subtract 3/5 from a whole fraction. </h3>

<u>\frac{5}{5} - \frac{3}{5} = \frac{2}{5}</u>

<h2>2/5 of your job remains to be done. </h2>

Hope this helps! :)

-Peredhel

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Player B’s longest home run distance is 500 ft. There are 5280 ft in 1 mi. How many home runs would Player B need to hit at his
Flura [38]
5280 ÷ 500 = (number of home runs needed to reach a mile distance)

5280 ÷ 500 = 10.56

cant hit part of a home run so round up to 11.
3 0
3 years ago
Round each number to the nearest million, thousand, and ten
yawa3891 [41]

Answer:

1)

<em> </em><em>a</em><em>.</em><em> </em>4,000,000

<em>b</em><em>.</em><em> </em>4,327,000

<em>c</em><em>.</em><em> </em>4,326,800

2)

<em>a</em><em>.</em><em> </em>30,000,000

<em>b</em><em>.</em> 29,727,000

<em>c</em><em>.</em><em> </em>29,726,520

3)

<em>a</em><em>.</em> 3,000,000

<em> </em><em>b</em><em>.</em><em> </em>2,867,000

<em>c</em><em>.</em> 2,867,000

Step-by-step explanation:

i mean,,

7 0
3 years ago
4r+3=-18 please help me
lorasvet [3.4K]
Subtract three from either side. That will give you 4r=-21. Divide -21 by 4 and r will equal -5.25.
6 0
3 years ago
A concentric-tube heat exchanger cools hot water owing at 1 kg/s from 80C to 30C. It uses air owing at 5 kg/s to perform this co
nordsb [41]

Answer:

The log-mean-temperature-difference is 24.03⁰C

Step-by-step explanation:

First we need to know if the heat exchanger is in parallel flow or counter-flow. However, counter flow arrangement is best used to recover heat.

L.M.T.D for counter flow is given as;

L.M.T.D =\frac{(T_h_f_1 -T_c_f_2)-(T_h_f_2 -T_c_f_1)}{2.3log[\frac{T_h_f_1 -T_c_f_2}{T_h_f_2 -T_c_f_1}]}

where;

Thf₁ is the initial temperature of the hot fluid = 80°C

Tcf₂ is the final temperature of the cold fluid = 51.5°C

Thf₁ - Tcf₂ = 80 - 51.5 = 28.5⁰C

Thf₂ is the final temperature of the hot fluid = 30°C

Tcf₁ is the initial temperature of the cold fluid = 10°C

Thf₂ - Tcf₁ = 30 - 10 = 20⁰C

L.M.T.D = \frac{28.5 -20}{2.3Log[\frac{28.5}{20}]} \\\\L.M.T.D = \frac{8.5}{0.3538} =24.03^oC

Therefore, the log-mean-temperature-difference is 24.03⁰C

3 0
3 years ago
Read 2 more answers
Stuck trying to solve this, didn’t pay attention :/
Gelneren [198K]

Answer:

Pay attention next time

6 0
2 years ago
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