All Answers 358129 Answers

Here's another way I like to do this kind of problem

Add

1/3 + 1/4 + 1/6 =

4/12 + 3/12 + 2/12 =

9/12     Now, take the reciprocal of this fraction 

12/9 =

4/3 =

1 + 1/3 hrs

We can write

log5832^1/3 =

(1/3)log 5832 =

(1/3)(3.765817515309918) =

1.255272505103306  ....... and we can write

10^{1.255272505103306}  = 18

x= 5.2x5.2

= 27.04

Great explanation  Ninja but I think you made an error.

(12.01) ÷ (6.022*10²³)  You have assumed the brackets which is not an unreasonable assumption

12.01 is the same thing as 1.201 x 10¹      

You would have been better off to have left it as 12.01 but it doesn't matter.

(1.201 x 10¹) ÷ (6.022 x 10²³)

(1.201 ÷  6.022)(10¹ ÷  10²³)

.1994354 x 10^-22     Good to here

this is the same as

(1.994354 x 10^-1) x 10^-22        It is 10^-1 NOT 10⁺¹

1.994354x 10^-23

Round down to 4 significant digits.

1.994 x 10^-23

An empty tank has a capacity of 55.2 litres.Water flows from a Tap A at 3.6 litres per mimute and 2.4 litres per minute from a Tap B.Tap A was turned on first.Tap B was turned on 2 minutes later.The taps were turned off at the same time when the tank was completely filled without overflowing.How much water flowed from Tap B ?

The tank after 2 min. has an empty capacity of$$55.2 - \dfrac{3.6\;l}{min}*2\;min=55.2\;l - 7.2\;l=48\;l$$

$$\left(
\dfrac{3.6\;l}{min}+\dfrac{2.4\;l}{min}
\right)
\times t_{min} = 48\;l \quad | \quad t_{min} = \mbox{time Tab B is running}$$

$$t_{min}=\dfrac{48\;l}{3.6\;l+2.4\;l}*min=\dfrac{48\;\not{l}}{6\;\not{l}}*min=8\;min$$

How much water flowed from Tap B ?

$$=\dfrac{2.4\;l}{min}*8\;min=\textcolor[rgb]{1,0,0}{19.2\;l}$$

So, assuming that 80 cm is the height of the first tank, the volume of water in the first tank =

40*20*80 = 64000cm^3

Note that the volume of water remains constant and is divided between the two tanks thusly:

40*20*h + 60*40*h = 64000   

800h + 2400h = 64000

3200h = 64000

 h = 20

So, both tanks would be filled to a height of 20cm

And the volume of water that would be poured from Tank A = 40*20*(80-20) =

48000cm^3

ninja is right except / is called a forward slash and that is what is used.

jboy314 has assumed that you meant this to be 3degrees.

When no units are mentioned it is usually assumed that the units are radians.

did you mean 3 degrees or 3 radians?

You have given a good answer Jboy314.  Thanks.

I intend to mention this question and your answer in my "End of Day Wrap" tonight.   

Why don't you take a look.   ♬ 

Log₃9^1/3 ?

$$\log_3{(9^\frac{1}{3})}=\log_3{(3^\frac{2}{3})}=\frac{2}{3}*\underbrace{\log_3{(3)}}_{=1}=\frac{2}{3}$$

An empty tank has a capacity of 55.2 litres.Water flows from a Tap A at 3.6 litres per mimute and 2.4 litres per minute from a Tap B.Tap A was turned on first.Tap B was turned on 2 minutes later.The taps were turned off at the same time when the tank was completely filled without overflowing.How much water flowed from Tap B?

When B is turned on, A has put 3.6(2) = 7.2 litres of water into the tank. So there remain 55.2 - 7.2 = 48 litres to fill.  So we have:

3.6T + 2.4T = 48     where T is the time it takes to fill the rest of the tank  .......Simpifying, we have

6T = 48

T  = 8 minutes

So, since B was on for 8 minutes, it put 2.4(8) -= 19.2  litres into the tank.

Very cute!

I have a copy of GeoGebra that I downloaded completely for free.  

I do my co-ordinate geometry graphing on Desmos

but

I draw all my euclidean geometry diagrams using GeoGebra.   It is well worth learning.

Yes Ninja, you left out a zero.  

But you can have my thumbs up anyway.

Ninja, I am not sure how well you thought this through.   It is correct but maybe there was a bit of luck involved here.  

"Since a 3/4 yard piece can't be cut out of a 1/3 yard-long piece, let's forget about this 1/3 yard piece"

This quote concerns me a little. There might be some left over from the 12 yard peice and together with the 1/3 yard maybe it would make another one.  Of course if you looked at the left over bit afterwards this would be alright.  That is probably what you were intending to do.

Otherwise it would be just

$$12\frac{1}{3}\div \frac{3}{4}$$

$${\frac{\left({\mathtt{12}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)}{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}} = {\frac{{\mathtt{148}}}{{\mathtt{9}}}} = {\mathtt{16.444\: \!444\: \!444\: \!444\: \!444\: \!4}}$$

Obviously you only want the whole bits so the answer is 16pieces

Anonymous posted: (2x + 1)(x + 3) = 2(x + 1/2)(x + 3) = (2x + 6)(x + 1/2).

Correct! This shows the two polynomials are the same (though note that the original question was to find the values of a, b and c - so there are two sets of answers to the original question).

If the volume of the tank is, say, V litres, then

A fills it at a rate of V/3 litres/hour

B fills it at a rate of V/4 litres/hour

C fills it at a rate of V/6 litres/hour

So A+B+C would fill it at a rate of V/3 + V/4 + V/6 = 3V/4 litres/hour

Time taken = V/(3V/4) = 4/3 hours  or 1hour 20 minutes. 

(2x + 1)(x + 3) = 2(x + 1/2)(x + 3) = (2x + 6)(x + 1/2).

$$\begin{array}{rcccccccl}
( 2x^4 &+& 3x^3 &-& 17x^2 &-& 27x &-& 9 ) : (x^2-2x-3) = 2x^2 +7x+3\\
-(2x^4&-&4x^3&-&6x^2)&&&&\\
--&-&-&-&-&&&&\\
&&7x^3&-&11x^2&-&27x&&\\
&-(&7x^3&-&14x^2&-&21x)&&\\
&&-&-&-&-&-&&\\
&&&&3x^2&-&6x&-&9\\
&&&&-(3x^2&-&6x&-&9)\\
&&&&&&&&0\\
\end{array}$$

$$\begin{array}{rcl}
2x^4 + 3x^3 -17x^2 -27x -9 &=& (x+1)(x-3) (ax+b)(x+c) \\
&=&(x+1)(x-3)(2x^2+7x+3)
\end{array}$$

$$\begin{array}{rcccrcc}
(ax+b)(x+c)&=&2x^2&+&7x&+&3 \\
(ax+b)(x+c)&=&ax^2&+&(ca+b)x&+&bc \\
\end{array} \\\\
\mbox{Coefficient comparison:} \\
a=2\\
ca+b=7\\
bc=3$$

$$\\2c+b=7 \qquad \Rightarrow \quad b=7-2c\\
bc=3\qquad \Rightarrow \quad (7-2c)c=3\qquad \Rightarrow \quad 2c^2-7c+3=0$$

$$c_{1,2}=\frac{7\pm\sqrt{49-4*2*3}}{2*2}=\frac{7\pm\sqrt{25}}{4}=\frac{7\pm5}{4}\\
c=c_1=\frac{12}{4}=3\\
c=c_2= \frac{2}{4}=\frac{1}{2}\\
b=b_1=7-2c_1=7-2*3=1\\
b=b_2=7-2c_2=7-2*\frac{1}{2}=6$$

1.

$$\\2x^4+3x^3-17x^2-27x-9=(x+1)(x-3)(2x+b_1)(x+c_1)=(x+1)(x-3)(2x+1)(x+3)$$

2.

$$\\2x^4+3x^3-17x^2-27x-9=(x+1)(x-3)(2x+b_2)(x+c_2)=(x+1)(x-3)(2x+6)(x+\frac{1}{2})$$

Well, if you insist on using grams, then at least use Graham’s number notation. Standard scientific notation will exceed the capacity of the servers.We do not want a crash.

Use the calculator on this site - it has a square root button.

You need to know the mass of the body with the fat.  Then express the percentage as a fraction and multiply by the mass of the body in question expressed in grams.

If you have an equation of a straight line in the form y = m*x + c where m and c are constants then the slope is given by m and the y-intercept is given by c. In this case m = 3, so that is the slope of your line.

Seems to me there are two sets of solutions here, unless I've screwed up somewhere (always a distinct possibility!). See below:

You still need to discuss the so called 'knock on effect'.

Suppose for example that you have the number 3.44445

Rounding to 4dp you get 3.4445, then to 3 dp 3.445, then to 2dp 3.45, then to 1dp 3.5, and finally to a whole number 4.

The 3.4445 is ok., but surely the others should be 3.444, 3.44, 3.4 and 3.

The implication would seem to be that when rounding, just the following digit should be considered, the ones after that ignored.