All Answers 352710 Answers

Thanks,

If you appreciate answers then you need to start saying so.

It is the ONLY payment any answerer ever gets here and it can mean a lot.

Jeff can get to any point in the square at (5,5), (5,-5), (-5,-5), (-5,5), so the answer is 100 points.

My diagram was only for the first part of the question.

You are right Dragan! Good work!

I corrected my ^_fuckup. 

GA

320 / 10 = 32 mph - his speed upstream

320 / 5   = 64 mph - his speed downstream

[64 + 32] / 2 = 48 mph - hid speed in still water.

[64 - 32] / 2 = 16 mph - the speed of the current .

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Hi there...

- Something tells me that you are being sarcastic...

NB!...: I'm serious...!!! 

Just look at this...: 
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Nr.1. (0.1)=(x) , x=(0.1) 
Nr.2. (0.01)=(x) , x=(0.01) 
Nr.3. (0.001)=(x) , x=(0.001) 
Nr.4. (0.0001)=(x) , x=(0.0001) 
Nr.5. (0.00001)=(x) , x=(0.00001) 
Nr.6. (0.000001)=(x) , x=(0.000001) 
Nr.7. (0.0000001)=(x) , x=(0.0000001) 
Nr.8. (0.00000001)=(x) , x=(0.00000001) 
Nr.9. (0.000000001)=(x) , x=(0.000000001) 
Nr.10. (0.0000000001)=(x) , x=(0.0000000001) 
Nr.11. (0.00000000001)=(x) , x=(0.00000000001) 
Nr.12. (0.000000000001)=(x) , x=(0.000000000001) 
Nr.13. (0.0000000000001)=(x) , x=(0.0000000000001) 
Nr.14. (0.00000000000001)=(x) , x=(0.00000000000001) <<--(The highest/lowest value before the calculator starts acting up.) 
Nr.15. (0.000000000000001)=(x) , x=(0.000000000000001000000000000001000000000000001000000000000001) 
Nr.16. (0.0000000000000001)=(x) , x=(0.0000000000000001) 
Nr.17. (0.00000000000000001)=(x) , x=(0) 
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Nr.1. (1.1)=(x) , x=(1.1) 
Nr.2. (1.01)=(x) , x=(1.01) 
Nr.3. (1.001)=(x) , x=(1.001) 
Nr.4. (1.0001)=(x) , x=(1.0001) 
Nr.5. (1.00001)=(x) , x=(1.00001) 
Nr.6. (1.000001)=(x) , x=(1.000001) 
Nr.7. (1.0000001)=(x) , x=(1.0000001) 
Nr.8. (1.00000001)=(x) , x=(1.00000001) <<--(The highest/lowest value before the calculator starts acting up.) 
Nr.9. (1.000000001)=(x) , x=(1.000000001000000083000006889000571787047458324939040969940401) 
Nr.10. (1.0000000001)=(x) , x=(1.000000000100000008280000685584056766359900254599741080858561) 
Nr.11. (1.00000000001)=(x) , x=(1.00000000001000000082750006847563066635843764116071480604915) 
Nr.12. (1.000000000001)=(x) , x=(1.000000000001000088900582611041856908252751101162755308435551) 
Nr.13. (1.0000000000001)=(x) , x=(1.000000000000099920072216268437233866881145005276001192933374) 
Nr.14. (1.00000000000001)=(x) , x=(1.00000000000000999200722162644436255341995847474320207467171) 
Nr.15. (1.000000000000001)=(x) , x=(1) 
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Nr.1. (10.1)=(x) , x=(10.1) 
Nr.2. (10.01)=(x) , x=(10.01) 
Nr.3. (10.001)=(x) , x=(10.001) 
Nr.4. (10.0001)=(x) , x=(10.0001) 
Nr.5. (10.00001)=(x) , x=(10.00001) 
Nr.6. (10.000001)=(x) , x=(10.000001) 
Nr.7. (10.0000001)=(x) , x=(10.0000001) <<--(The highest/lowest value before the calculator starts acting up.) 
Nr.8. (10.00000001)=(x) , x=(10.000000010000000900000081000007290000656100059049005314410478) 
Nr.9. (10.000000001)=(x) , x=(10.000000001000000083000006889000571787047458324939040969940401) 
Nr.10. (10.0000000001)=(x) , x=(10.000000000100000008280000685584056766359900254599741080858561) 
Nr.11. (10.00000000001)=(x) , x=(10.00000000000999911264904615907617648764224245910671876622973) 
Nr.12. (10.000000000001)=(x) , x=(10.000000000001000088900582611041856908252751101162755308435551) 
Nr.13. (10.0000000000001)=(x) , x=(10.000000000000099475983006416853299441709647604154773422567477) 
Nr.14. (10.00000000000001)=(x) , x=(10) 
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Nr.1. (100.1)=(x) , x=(100.1) 
Nr.2. (100.01)=(x) , x=(100.01) 
Nr.3. (100.001)=(x) , x=(100.001) 
Nr.4. (100.0001)=(x) , x=(100.0001) 
Nr.5. (100.00001)=(x) , x=(100.00001) 
Nr.6. (100.000001)=(x) , x=(100.000001) 
Nr.7. (100.0000001)=(x) , x=(100.0000001) <<--(The highest/lowest value before the calculator starts acting up.) 
Nr.8. (100.00000001)=(x) , x=(100.000000009999993800003843997616721477632683867736002003678758) 
Nr.9. (100.000000001)=(x) , x=(100.000000001000003636013220544069898238149993913377869041931836) 
Nr.10. (100.0000000001)=(x) , x=(100.000000000100001784631848539969044287564355873494918390313594) 
Nr.11. (100.00000000001)=(x) , x=(100.000000000010004441719511309603803521267403815292043331114921) 
Nr.12. (100.000000000001)=(x) , x=(100.000000000000994759830064762261266123026921032269075120175044) 
Nr.13. (100.0000000000001)=(x) , x=(100) 
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Nr.1. (1000.1)=(x) , x=(1000.1) 
Nr.2. (1000.01)=(x) , x=(1000.01) 
Nr.3. (1000.001)=(x) , x=(1000.001) 
Nr.4. (1000.0001)=(x) , x=(1000.0001) 
Nr.5. (1000.00001)=(x) , x=(1000.00001) 
Nr.6. (1000.000001)=(x) , x=(1000.000001) <<--(The highest/lowest value before the calculator starts acting up.) 
Nr.7. (1000.0000001)=(x) , x=(1000.000000099999970000008999997300000809999757000072899978130007) 
Nr.8. (1000.00000001)=(x) , x=(1000.000000010000008000006400005120004096003276802621442097153678) 
Nr.9. (1000.000000001)=(x) , x=(1000.000000000999989425111829442403646581437401299481257985696801) 
Nr.10. (1000.0000000001)=(x) , x=(1000.000000000100044417200117538744240466412227275322115681245158) 
Nr.11. (1000.00000000001)=(x) , x=(1000.000000000010004441719511309603803521267403815292043331114921) 
Nr.12. (1000.000000000001)=(x) , x=(1000) 
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Nr.1. (10000.1)=(x) , x=(10000.1) 
Nr.2. (10000.01)=(x) , x=(10000.01) 
Nr.3. (10000.001)=(x) , x=(10000.001) 
Nr.4. (10000.0001)=(x) , x=(10000.0001) 
Nr.5. (10000.00001)=(x) , x=(10000.00001) <<--(The highest/lowest value before the calculator starts acting up.) 
Nr.6. (10000.000001)=(x) , x=(10000.000001000001000001000001000001000001000001000001000001000001) 
Nr.7. (10000.0000001)=(x) , x=(10000.00000010000077000592904565365153311680499939849536841433679) 
Nr.8. (10000.00000001)=(x) , x=(10000.000000010000803764598560786330397374036951349779981817138643) 
Nr.9. (10000.000000001)=(x) , x=(10000.000000001000444172201353095740681469055223792983489844903891) 
Nr.10. (10000.0000000001)=(x) , x=(10000.000000000100044417200117538744240466412227275322115681245158) 
Nr.11. (10000.00000000001)=(x) , x=(10000) 
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Nr.1. (100000.1)=(x) , x=(100000.1) 
Nr.2. (100000.01)=(x) , x=(100000.01) 
Nr.3. (100000.001)=(x) , x=(100000.001) 
Nr.4. (100000.0001)=(x) , x=(100000.0001) <<--(The highest/lowest value before the calculator starts acting up.) 
Nr.5. (100000.00001)=(x) , x=(100000.000010000100001000010000100001000010000100001000010000100001) 
Nr.6. (100000.000001)=(x) , x=(100000.000000999994000035999784001295992224046655720065679605922364) 
Nr.7. (100000.0000001)=(x) , x=(100000.00000010000077000592904565365153311680499939849536841433679) 
Nr.8. (100000.00000001)=(x) , x=(100000.000000009997165803494709249927645512497207041803648665603301) 
Nr.9. (100000.000000001)=(x) , x=(100000.00000000100408215121026364340677061330075796304042883063453) 
Nr.10. (100000.0000000001)=(x) , x=(100000) 
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Nr.1. (1000000.1)=(x) , x=(1000000.1) 
Nr.2. (1000000.01)=(x) , x=(1000000.01) 
Nr.3. (1000000.001)=(x) , x=(1000000.001) 
Nr.4. (1000000.0001)=(x) , x=(1000000.0001) 
Nr.5. (1000000.00001)=(x) , x=(1000000.00001) <<--(The highest/lowest value before the calculator starts acting up.) 
Nr.6. (1000000.000001)=(x) , x=(1000000.000001000008000064000512004096032768262146097168777350218802) 
Nr.7. (1000000.0000001)=(x) , x=(1000000.00000010000077000592904565365153311680499939849536841433679) 
Nr.8. (1000000.00000001)=(x) , x=(1000000.000000010011717714412948915410596562216418095739253221970995) 
Nr.9. (1000000.000000001)=(x) , x=(1000000) 
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Nr.1. (10000000.1)=(x) , x=(10000000.1) 
Nr.2. (10000000.01)=(x) , x=(10000000.01) 
Nr.3. (10000000.001)=(x) , x=(10000000.001) 
Nr.4. (10000000.0001)=(x) , x=(10000000.0001) <<--(The highest/lowest value before the calculator starts acting up.) 
Nr.5. (10000000.00001)=(x) , x=(10000000.000010000600036002160129607776466587995279716783006980418825) 
Nr.6. (10000000.000001)=(x) , x=(10000000.000001000241058095000895215746995025801218093560548092090194) 
Nr.7. (10000000.0000001)=(x) , x=(10000000.000000100582847425979322379391862022863889379789063687349082) 
Nr.8. (10000000.00000001)=(x) , x=(10000000) 
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Nr.1. (100000000.1)=(x) , x=(100000000.1) 
Nr.2. (100000000.01)=(x) , x=(100000000.01) 
Nr.3. (100000000.001)=(x) , x=(100000000.001) <<--(The highest/lowest value before the calculator starts acting up.) 
Nr.4. (100000000.0001)=(x) , x=(100000000.000100010001000100010001000100010001000100010001000100010001) 
Nr.5. (100000000.00001)=(x) , x=(100000000.000009998700168978032855728755261815963924689790327257456531) 
Nr.6. (100000000.000001)=(x) , x=(100000000.000000998378633099845850339049383800707650775141170738720318) 
Nr.7. (100000000.0000001)=(x) , x=(100000000) 
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Nr.1. (1000000000.1)=(x) , x=(1000000000.1) 
Nr.2. (1000000000.01)=(x) , x=(1000000000.01) <<--(The highest/lowest value before the calculator starts acting up.) 
Nr.3. (1000000000.001)=(x) , x=(1000000000.001001001001001001001001001001001001001001001001001001001001) 
Nr.4. (1000000000.0001)=(x) , x=(1000000000.000100020004000800160032006401280256051210242048409681936387) 
Nr.5. (1000000000.00001)=(x) , x=(1000000000.000010013618521188816790835536329407994873027317151325803092) 
Nr.6. (1000000000.000001)=(x) , x=(1000000000) 
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Nr.1. (10000000000.1)=(x) , x=(10000000000.1) 
Nr.2. (10000000000.01)=(x) , x=(10000000000.01) 
Nr.3. (10000000000.001)=(x) , x=(10000000000.001) <<--(The highest/lowest value before the calculator starts acting up.) 
Nr.4. (10000000000.0001)=(x) , x=(10000000000.000099186669311644514977187066058321761555246974806585994842) 
Nr.5. (10000000000.00001)=(x) , x=(10000000000) 
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Nr.1. (100000000000.1)=(x) , x=(100000000000.1) 
Nr.2. (100000000000.01)=(x) , x=(100000000000.01) <<--(The highest/lowest value before the calculator starts acting up.) 
Nr.3. (100000000000.001)=(x) , x=(100000000000.001008064516129032258064516129032258064516129032258064516129) 
Nr.4. (100000000000.0001)=(x) , x=(100000000000) 
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Nr.1. (1000000000000.1)=(x) , x=(1000000000000.1) <<--(The highest/lowest value before the calculator starts acting up.) 
Nr.2. (1000000000000.01)=(x) , x=(1000000000000.010101010101010101010101010101010101010101010101010101010101) 
Nr.3. (1000000000000.001)=(x) , x=(1000000000000) 
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Nr.1. (10000000000000.1)=(x) , x=(10000000000000.1) <<--(The highest/lowest value before the calculator starts acting up.) 
Nr.2. (10000000000000.01)=(x) , x=(10000000000000) 
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Nr.1. (100000000000000.1)=(x) , x=(100000000000000) 
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There are three important aspects of a calculator/instrument...: 
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Nr.1. "Resolution" = A way to display results accurately with the amount of digits required for the specific situation, and with as little rounding off as possible.
Nr.2. "Accuracy" = How close to the absolute value, the result is for a given calculation/measurement. Closely related to the resolution mentioned above.
Nr.3. "Precision" = The repeatability of accuracy.
---------------------------------------------------------------------------------------------------------------

This calculator doesn't just struggle with resolution and accuracy.
But the calculator even has trouble with "Precision".
Especially when working with decimals...
Whole-numbers, or Natural-numbers, seem to be unaffected, but I might be wrong.

As you can clearly see something is going on with this calculator which makes me question the validity of the results it spits out.

So please spare me your sarcasm...

As an example...: 
I've seen many times online that people define "1 inch equals 2.5cm" which at first glance doesn't seem that bad, but in reality equals an error of "25.344000 Meters pr. international land-mile".
That's a pretty big difference, so way less accurate in comparison.
 

Kind regards

BizzyX

-----

I tried both 3360 and 2730, and both are incorrect. Help!

The angles are 14 degrees and 76 degrees.

If Elena is on page 55 and has read 20 more pages than Andre, it means Andre is on page 35.

If we let x be the number of pages in the book, we can express this as x*0.2=35.  In other words 20% of x is 35.

If x*0.2=35 then x=35/0.2

Cheers

Let x be the number of coach tickets bought. Since we have 8 the number of people who gets first class is 8 minus the number of coach tickets, or 8-x.

With that in mind you can write the equation as 290*x+980*(8-x)=5080.

Factoring gives:

    290x+980*8-980x=5080

    290x-980x+7840=5080

Move the x's to one side and the numbers to the other

980x-290x=7840-5080

690x=2760

x=2760/690

x=4

So that gives 4 coach tickets and 4 first class.

You can double check by 4*980 + 4*290 =5080
 

cheers

Using the angle bisector theorem, we can work out that AB = 25/3 and BF = 31/11.  You can then use the formula for an angle bisector to find BE:  https://proofwiki.org/wiki/Length_of_Angle_Bisector

in other words, dividing by fractions means multiplying by the fraction's reciprocal (flip the numerator and denominator around)
so, 3 / (2/7)= 3 * (7/2)
Multiply the numerators and denominators together (3=3/1):
(3*7)/2

you get 21/2

a ÷ ( b / c ) = a * ( c / b )

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-

-

In this case, a = 3, b = 2, and c = 7.

-

-

✧･ﾟ: *✧･ﾟ:*  I hope this helped!  *:･ﾟ✧*:･ﾟ✧

Jackies number is x and Adrians is y.
 

y=x-6

x*(x-6)=567

factor out

x^2+6x=567

Set right side to zero

x^2+6x-567=0

Now you want to use AC method and find the two factors that have a difference of 6 and product of 567 (in this case 27 and -21)
Then you can rewrite the equation as x^2+27x-21x-567=0

Factor this out to get x(x+27)-21(x+27)

x is then 27 or -21  and y is 21 or -27

Edit: Missunderstood question, corrected.

Let x be the large number and y the smaller number. we know that 2*x+2*y =310 and that x-y=55 which can also be written as y=x-55

2x+2y=310

Divide by 2

x+y=155

Since y=x-55 we can substitute y for this and get

x+x-55=155

2x=155+55

2x=210

x= 105

y= 105-55 => y=50

Check by 105*2+50*2 => 210+100=310

cheers

The quick equation for this is as follows:

x=850/(2*6*24)

x=850/288

x=2.95, rounded up to x=3

Basically you calculate how many golf balls are in a case (24*6) and then multiply this by the price per golf ball which gives you the price per case.
Since you know they need to raise 850 you simply divide it by the value per case, the reverse of price*quantity=revenue and you have the number of cases.

Hope it helps,
Cheers

I'm sorry you felt that way. I really do appreaciate all answers everyone teaches me. Many of you guys have helped me a lot, and yes, I will change the latex.

Suppose that you have an enormous grapefruit that is 92% water (by weight). The grapefruit weights 100 pounds. If the water content of the grapefruit evaporates until it is 90% water (by weight), then approximately how much does the grapefruit now weigh?

Sorry to say,but you're wrong!

You were correct in this:>   https://web2.0calc.com/questions/help-asap-thanks_2#r5

I'm gonna use the same calculations for this math problem.

[ 0.92 * 100 - 0.90 (100 - x)] = x      ( "x" represents the weight loss)

92 - ( 90 - 0.90x) = x

92 - 90 + 0.90x = x

2 + 0.90x = x

2 + 0.90x - 0.90x = x - 0.90x

2 = 0.1x

x = 20

New weight is    100 - x = 80  

The sum of angles  J  and  M  will be  180°.

J + M  =  (23x + 4) + (12x + 15)  =  180

                                   35x + 19  =  180

                                            35x  =  161

                                               x  =  4.6

Angle K will have the same size as angle M:  12(4.6) + 15  =  70.2°

If you subract the persons weight by the maximum capacity of the elevator you have the total weight which can be "filled" by books. If you then divide this number by the weight of a box of books you'll have the number of boxes of books he could bring.

let x be number of books.

x=(920-170)/40

x=750/40

x= 18.75 boxes.

Techincally that means he'll probably only bring 18 boxes or 18 boxes and one which is 3/4 filled with books.

Cheers.

Let A be the number of adult tickets and C be the amount of children tickets. We can form the equation 105A + 60C = 7155, as we converted it from dollars to regular integers. We can simplify the equation by dividing by 5, giving 21A + 12C = 1431. Dividing by 3 gives 7A + 4C = 477. 477 divided by 7 gives 68 remainders 1, and if we remove one more seven and add it to one, we will have a multiple of 4. This means 67 adult tickets and 2 children tickets.

If it is filled to half its capacity when placed on end, it will also be filled to half its capacity when placed on its side.

Since the height of the water is 4 feet when it is filled to half its capacity, the height of the tank is 8 feet.

I'm assuming that when the tank contains 36pi cubic feet of water, this only fills half the tank; therefore, the tank will hold 72pi cubic feet of water when completely filled.

Sice the volume of the tank is 72pi and the formula for volume is    V  =  pi·r²·h

                                                                                                      72pi  =  pi·r²·8

Divide both sides by 8pi:                                                                    9  =  r²

The radius of the tank is 3 feet.and the diameter is 6 feet.

Half-ways, the water will reach 3 feet.

C'mon man. First, at least simplify the latex. Also, it's literally rounding. If you do not know how to find the answer to a problem like this, then you need to learn how to, not just ask to get an answer to your AoPS thing or whatever.