Question

please simplify -4/9 - (-6a/9) when a=-1/4

Answer 1

-4/9 -(6a/9)
-4 -6a
---------
9
for a=-1/4

-4 -6(-1/4) -10/4
----------- = ------ =-5/18
9 9

Answer 2

@radar my answer is -11/18
-10 1/4 become -5 1/2 then change to improper fraction -11/2 x 1/9
pls check my answer if you don't mind.tnx

Answer 3

Let me see......ah so!, I think I booed booed.

-4/9 - (-6a)/9= -4/9 +6a/9 Note the sign difference.
substituting -1/4 for a, then would give:
-4/9 + 6(-1/4)/9
-4/9 + (-6/4)/9
-4/9 +(-6/4)(1/9)
-4/9 -1/6
-8/18 -3/18=-11/18

wifeypearl, you my dear are absolutely correct. sorry about my carelessness.

Answer 4

@radar: thank you so much for your time...have 1 more problem unsolved: kindly solve & show your solution. pedro is 30y/o & michael is 24y/o.how many years ago was pedro's age 11/2 times michael's age? please help

Answer 5

Let x = the number of years ago when Pedro's age was 11/2 times Michael's age. ( Is that one and one half or eleven halves?) I'll work it both ways.
30-x is Pedro's age back when.
24-x is Michael's age back when.

Answer 6

eleven halves

Answer 7

30-x 3
----- = ---- Doing the 1 1/2 or 3/2 first.
24-x 2
cross multiplying
60-2x=72-3x
x=72-60
x=12 or 12 years ago. Pedro was 18 and Michael was 12 and Pedro was 1 1/2 times old as Michael.

I am thinking that this is the way the problem is meant to b.

Answer 8

don't have any idea 'bout this but i'l try to understand.thank u once again

Answer 9

However, we will do for the 11/2 (I seriously doubt this fraction as it will give some odd ball results!

30-x 11
----- = ----
24-x 2

Cross multiplying getting:

60-2x=264-11x
9x=204
x= 22.666667 years so you see why I doubt the 11/2 anyway
Pedro would of been 30-22.66667=7.3333333 yrs old
Michael would of been 24-22.666666 =1.33333 yrs old

Answer 10

Review the problem and verify.

Answer 11

not ration but ratio/proportion

Answer 12

I am sure the answer is 12 years ago.

Answer 13

yeah.i'll confirm tomorrow...tomorrow? it's 12:13am now in our place.bye for now

Answer 14

Good luck in your studies.

Answer 15

@radar:hi

Answer 16

have 4 more probs. to solve

Answer 17

You are busy wifeypearl

Answer 18

Where is your problem posted?

Answer 19

@radar:rita's age is 3 times anna's. anna is twice as old as she is now, rita's age will only be twice anna's. how old is each now?

Answer 20

carla plans to wire three identical light fixtures together with a 420W appliance & 200W appliance. if the circuit is 110V & has 10 amp fuse, what are the highest voltage bulbs that she can use for each fixture? use the formula: total wattage = voltage x ampere

Answer 21

Both of these problems have some confusing issues. The first one states that Rita's age is 3 times Anna's, and in the next breath it states, Anna is twice as old as she is now. ???
You get my point, which is it? Is it Rita 3 times anna or Anna twice as old as Rita?

The second question, I believe the question should be asking "What are the highest WATTAGE bulbs that she can use?

Please review your material and verify.

Answer 22

I am going to attempt the second problem with what I think it is asking. "Carla plans to wire three identical light fixtures with a 420W applicance and a 200W applicance. If the circjuit is 110 volt and has 10 amp fuse, what are the hightest Wattage bulbs that she can use for each fixture (and still continue to use the 10 amp fuse)?"

Step 1. Determine the capacity (watts) of the circuit that she intends to make the connections. Watts= voltage X current, 110 X 10 = 1100 Watts. She cannot exceed that or the fuse will "blow"

Step 2. Sum up the wattage of the appliances: 420+200=620 watts.

Step 3. Determine how many watts of load (lights) can be added.
1100 watts - 620 watts = 480 watts available.

Step 4. Determine wattage of the three identical light fixtures: 480 watts divided by 3
480/3=160 watts. That is the answer. Their voltage will be what the circuit provides or 110 volt bulbs. I don't think that is what the problem wants, as that would not require any calculation lol!.

Answer 23

sorry, it's only now i've opened my pc...tnx a lot for helping me. i've got the same answer with you regarding carla but the age problem - not yet solved. i've already asked the help of my daughter's math teacher for RITA but she was also confused to the problem: AS OLD AS SHE IS NOW... it's hard to determine the word SHE, if it is rita or anna...these problems were copied from the book given by my son's teacher.thank you so much for being nice & patient.

Answer 24

O.K, no problem wifeypearl. I think the age problem may be similar to the other one where so many years ago she was three times and now she is only twice blah blah. But we do need the information to solve it.

Answer 25

well bye for now. i'l try to solve 1 last given problem now & if ever can't get the right equation, i'll send it you tomorrow, again hehehe!

Answer 26

O.K. I will stand by for a few minutes.

Answer 27

all problems are solved except RITA. hope you don't mind if i'll try others opinion about it...hope against hope- hoping somebody will get the equation by chance...huhuhu

Answer 28

Yes post it and we will see what happens. Good luck with it.

Answer 29

hi radar. i've got the equation for the problem rita. it was solved by my two friends (doesn't know each other). their answer is in variable since there is no specified given number. answer is:
anna's age now ---- = x ---------- 2x anna's
rita's age ---- = 3x --------- 2(2x)

nobody answered to my posted problem

Answer 30

Glad you and your friends got it worked out.