anonymous
  • anonymous
I'll give you a medal and be a fan if you HELP PLEASE 1. The pump used to fill the water tank on a fire truck takes 40 minutes to fill the tank. If the pump sending water to the hose is on, it takes 60 minutes to fill the tank. If the pump filling the water tank is off, how long does the pump sending water to the hose take to empty the tank? A. 120 minutes B. 60 minutes C. 30 minutes D. 24 minutes 2. Annmarie can plow a field in 240 minutes. Gladys can plow a field 80 minutes faster. If they work together, how many minutes does it take them to plow the field? A. 96 minutes B. 160 minutes C. 400 minutes D. 480 minutes 3.Brody can fill a bowl with candy in 3 minutes. While Brody fills the bowl, Hudson takes the candy out of the bowl. With Hudson taking candy out of the bowl, it takes 5 minutes for Brody to fill the bowl. Which of the following can be used to determine the amount of time it takes for Hudson to empty the bowl if Perry does not add candy? A. 1 over 3 minus 1 over x equals 1 over 5 B. 1 over 3 minus 1 over 5 equals 1 over x C. 1 over 5 minus 1 over x equals 1 over 3 D. 1 over 3 minus 1 over x equals x over 5 4. Leanne can clean a fish tank in 30 minutes. Karl can clean a fish tank in 20 minutes. If they work together, how many minutes does it take them to clean a fish tank? A. 6 minutes B.10 minutes C. 12 minutes D. 50 minutes 5.Genna can build a cabinet in 4 hours. Claude can build a cabinet in only 3 hours. Which of the following can be used to determine the amount of time it would take for Genna and Claude to build a cabinet together? A. 1 over 4 plus 1 over x equals 1 over 3 B. 1 over 4 plus 1 over 3 equals 1 over x C. 1 over x plus 1 over 3 equals 1 over 4 D. 1 over 3 plus 1 over 4 equals x over 7
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@ganeshie8
anonymous
  • anonymous
@iGreen
anonymous
  • anonymous
@Michele_Laino

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
@TheSmartOne
Michele_Laino
  • Michele_Laino
Question 5) the working rate of Genna is W/4, the working rate of Claude is W/3, where W is the work to do. When Gwenna and Claude work together, then the working rate is: W/4 + W/3 so we can write: \[\Large \left( {\frac{W}{4} + \frac{W}{3}} \right)t = W\] where t is time needed to Genna and Claude to do the work W Please solve that equation for T, what do you get?
Michele_Laino
  • Michele_Laino
oops.. for t, what do you get?
anonymous
  • anonymous
3 hrs?
Michele_Laino
  • Michele_Laino
hint: what is: \[\frac{1}{4} + \frac{1}{3} = ...?\]
anonymous
  • anonymous
\[\frac{ 7 }{ 12 }\]
Michele_Laino
  • Michele_Laino
ok! so, from preceding equation, we can write: \[\frac{{7W}}{{12}}t = W\] or: \[\frac{7}{{12}}t = 1\]
anonymous
  • anonymous
ok
Michele_Laino
  • Michele_Laino
namely: \[t = \frac{{12}}{7}\]
anonymous
  • anonymous
so the answer is D then?
Michele_Laino
  • Michele_Laino
yes!
anonymous
  • anonymous
Thanks ! can you help with the others ?
Michele_Laino
  • Michele_Laino
ok! Question 4)
Michele_Laino
  • Michele_Laino
the working rate of Leanne is W/30, where the working rate of Karl is W/20, where as usual, W is the work to do. So when Leanne and Karl work together the working rate is: W/30+ W/20, and we can write: \[\Large \left( {\frac{W}{{30}} + \frac{W}{{20}}} \right)t = W\] where t is time neeeded to Leanne and Karl to finish the work W. Please solve for t
Michele_Laino
  • Michele_Laino
oops...whereas the working rate of Karl
Michele_Laino
  • Michele_Laino
as before, we can write: \[\Large \left( {\frac{1}{{30}} + \frac{1}{{20}}} \right)t = 1\]
Michele_Laino
  • Michele_Laino
what is t?
anonymous
  • anonymous
\[\frac{ 5 }{ 15 } ?\]
Michele_Laino
  • Michele_Laino
we have: \[\frac{1}{{30}} + \frac{1}{{20}} = \frac{{2 + 3}}{{60}} = ...?\]
anonymous
  • anonymous
i entered the wrong thing my bad!
anonymous
  • anonymous
\[\frac{ 1 }{ 12 }\]
Michele_Laino
  • Michele_Laino
ok! So we have: \[\frac{t}{{12}} = 1\]
Michele_Laino
  • Michele_Laino
and therefore: \[t = 12 \times 1 = ...?\]
anonymous
  • anonymous
12?
Michele_Laino
  • Michele_Laino
that's right!
Michele_Laino
  • Michele_Laino
it is 12 minutes
anonymous
  • anonymous
Ok Thanks!
anonymous
  • anonymous
Have time for another one?
Michele_Laino
  • Michele_Laino
yes!
Michele_Laino
  • Michele_Laino
question 3)
anonymous
  • anonymous
ok
Michele_Laino
  • Michele_Laino
I'm pondering....
Michele_Laino
  • Michele_Laino
the working rate of Brody is: W/3, when he works alone whereas the working rate of Brody is W/5, when Brody and Hudson work together W is the work to be done Now the working rate of Hudson is: W/x, where x is the unknown quantity. So, we can write: \[\Large \frac{W}{3} - \frac{W}{x} = \frac{W}{5}\] or: \[\Large \frac{1}{3} - \frac{1}{x} = \frac{1}{5}\] so, what is the right option?
Michele_Laino
  • Michele_Laino
hint: \[\Large \frac{1}{x} = \frac{1}{3} - \frac{1}{5}\]
Michele_Laino
  • Michele_Laino
@amielli
anonymous
  • anonymous
2/15
Michele_Laino
  • Michele_Laino
yes! What is the right option?
anonymous
  • anonymous
D?
Michele_Laino
  • Michele_Laino
I think that option D is wrong
Michele_Laino
  • Michele_Laino
please look at this equation: \[\Large \frac{1}{3} - \frac{1}{x} = \frac{1}{5}\] do you recognize to which option corresponds to?
anonymous
  • anonymous
yes that would be A
Michele_Laino
  • Michele_Laino
yes! correct!
anonymous
  • anonymous
I dont get it though
Michele_Laino
  • Michele_Laino
since Brody is filling the bowl, he is adding candies to that bowl. Hudson is subtracting candies to that bowl. Here is why I used the minus sign in the subsequent equation: \[\large \frac{W}{3} - \frac{W}{x} = \frac{W}{5}\]
Michele_Laino
  • Michele_Laino
now, is it clear?
anonymous
  • anonymous
Oh yes I do
anonymous
  • anonymous
Thanks
Michele_Laino
  • Michele_Laino
ok!
Michele_Laino
  • Michele_Laino
question 2)
Michele_Laino
  • Michele_Laino
here we can write: working rate of Annmarie is W/240, whereas the working rate of Gladies is W/80. Now when Annmarie and Gladies work together, the working rate is: W/240 + W/80. As usual W is the work to be done. So we can write: \[\Large \left( {\frac{W}{{240}} + \frac{W}{{80}}} \right)t = W\] or: \[\Large \left( {\frac{1}{{240}} + \frac{1}{{80}}} \right)t = 1\] Please solve for t
Michele_Laino
  • Michele_Laino
hint: \[\Large \frac{1}{{240}} + \frac{1}{{80}} = \frac{{1 + 3}}{{240}} = ...?\]
Michele_Laino
  • Michele_Laino
please wait a moment, I think to have made an error
anonymous
  • anonymous
ok
Michele_Laino
  • Michele_Laino
the working rate of Gladies is: \[\Large \frac{W}{{240 - 80}} = \frac{W}{{160}}\]
Michele_Laino
  • Michele_Laino
so the right equation is: \[\Large \left( {\frac{W}{{240}} + \frac{W}{{240 - 80}}} \right)t = W\]
Michele_Laino
  • Michele_Laino
or: \[\Large \left( {\frac{1}{{240}} + \frac{1}{{160}}} \right)t = 1\]
Michele_Laino
  • Michele_Laino
please solve for t, keep in mind that: \[\Large \frac{1}{{240}} + \frac{1}{{160}} = \frac{{2 + 3}}{{480}} = \frac{5}{{480}}\]
Michele_Laino
  • Michele_Laino
after a substitution, we have: \[\Large \frac{5}{{480}}t = 1\] what is t?
anonymous
  • anonymous
just simplify it?
Michele_Laino
  • Michele_Laino
hint: \[\Large t = \frac{{480}}{5} = ...?\]
anonymous
  • anonymous
96
anonymous
  • anonymous
I get that one
anonymous
  • anonymous
So the last one
Michele_Laino
  • Michele_Laino
ok! question 1)
Michele_Laino
  • Michele_Laino
this question is similar to the question of candies, namely question 3). So the rate of working for the pump is W/40, when the pump works alone the working rate of the firehose is W/x, where x is the unknown quantity and W/60 is the working rate of the pump, when the pump and the firehose work together. So we can write: \[\Large \frac{W}{{40}} - \frac{W}{x} = \frac{W}{{60}}\] or: \[\Large \frac{1}{{40}} - \frac{1}{x} = \frac{1}{{60}}\]
anonymous
  • anonymous
so the answer is 60?
Michele_Laino
  • Michele_Laino
no, since we have to compute the value of x
Michele_Laino
  • Michele_Laino
hint: \[\Large \frac{1}{x} = \frac{1}{{40}} - \frac{1}{{60}} = \frac{{3 - 2}}{{120}} = ...\]
Michele_Laino
  • Michele_Laino
please continue my computation
anonymous
  • anonymous
1/120
Michele_Laino
  • Michele_Laino
ok! so we have: \[\Large x = \frac{{120}}{1} = ...{\text{minutes}}\]
anonymous
  • anonymous
thanks very much I got 80%
anonymous
  • anonymous
number 5 was wrong
Michele_Laino
  • Michele_Laino
are you sure?
Michele_Laino
  • Michele_Laino
I'm very sorry, the correct answer is option B. Sorry again
anonymous
  • anonymous
That is totally fine
anonymous
  • anonymous
Thanks!
Michele_Laino
  • Michele_Laino
since we can write this: \[\Large \left( {\frac{W}{4} + \frac{W}{3}} \right)t = W\] and then: \[\Large \frac{1}{4} + \frac{1}{3} = \frac{1}{t}\] Sorry again
Michele_Laino
  • Michele_Laino
thanks for your comprehension!
anonymous
  • anonymous
Thanks for helping me

Looking for something else?

Not the answer you are looking for? Search for more explanations.