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Unfortunately, a while back, there was a Vine created a boy unfortunate enough not to receive a good education. He said that 9+10=21, and was teased nationwide. He later ran away.

If you ever see these posts in this forum, just ignore them, as the posters are probably just trolls.

P.S. I assure you that 9+10 DOES equal 19.

For questions like these, you can use the calculator on this site by going to the top of the page and clicking the "Home" tab, but I'll help you out here.

First, follow PEMDAS (Parenthases, Exponents, Division/Multiplication, Addition/Subtraction)

This first step in this equation is 2000*4000, as this is multiplication, which comes before subtraction. 2000*4000=8,000,000 (8 million).

Next, do subtraction. 8,000,000-60,000,000,000=-59,992,000,000.

In conclusion, 2000*4000-60000000000=-59992000000.

The formula for finding the area of a circle is pi*r^2. First, find the radius of the circle. As an example, let's say that the radius of the circle is 4 (if only a diameter is given, divide the diameter by 2 points o find the radius). Then, do 4^2 (4 squared), which is 16. Then, multiply r^2 by pi (approximately 3.14, but to be specific, pi goes infinitely) 16*3.14 is 50.24, which means the area of a circle with a radius of 4 is approximately 50.24.

P.S. If you have a calculator, there may be a pi symbol (π), which gives you a more accurate value for pi. If there isn't, but you still want to be specific, you can use what I remember from a few years ago (3.14159265358979323)

I'm not sure I understand, but I'll try to help you out here...

If you mean 11.72/237.4*100, then do 11.72/237.4. That gets you approximately 0.0493.

Then, multiply that by 100. This gets you 4.93. The answer to 11.72/237.4*100 is approximately 4.93.

However, if you mean 11.72/(237.4x/100), it is a bit different. First, solve what's in the parentheses. 237.4x/100 is just the same equation without x, and then adding x back at the end. 237.4/100 is 2.374. This means the answer to that is 2.374x. Next, you must multiply 2.374x by 11.72. Multiply the two numbers without x, and then add x back at the end. 2.374*11.72 is 27.82 (when rounding to the nearest hundredth). Add x and you have 27.82x. In conclusion, 11.72(237.4x/100) is approximately 27.82x.

YOU CANNOT APPLY THE QUADRATIC FORMULA TO THAT BECAUSE IT IS NOT AN EQUATION.

IT MUST = 0   !!

Let   M=2000,     G=1.03      R=1.04

Let P be the present value

\(Original\; value    =  P\\ End\; of \;1\; year    =  RP-M\\ End\; of \;2nd\; year =  R(RP-M) - GM\\ End\; of \;2nd\; year =  R^2P- RM - GM\\ End\; of \;2nd\; year = R^2P-M(R+G) \\ End\; of \;3rd\; year =  R(R^2P- M( R+G)) - G^2M\\ End\; of \;3rd\; year =  R^3P- M( R^2+GR) - G^2M\\ End\; of \;3rd\; year =  R^3P- M( R^2+GR+G^2) \\ \)

\(End\; of \;4th\; year =  R(R^3P- M( R^2+GR+G^2))-G^3M \\ End\; of \;4th\; year =  R^4P- M( R^3+GR^2+G^2R)-G^3M \\ End\; of \;4th\; year =  R^4P- M( R^3+GR^2+G^2R+G^3) \\ End\; of \;nth\; year =  R^nP- M( R^{n-1}+R^{n-2}G^1+.......R^1G^{n-2}+G^{n-1}) \\ End\; of \;nth\; year =  R^nP- M*\displaystyle\sum_{k=0}^{n-1}\;R^{n-1-k}G^k\)

NOW for our question the amont at the end of the 20th year =0

\(End\; of \;20th\; year = 1.04^{20}P- 2000*\displaystyle\sum_{k=0}^{19}\;1.04^{19-k}*1.03^k\\ 0 = 1.04^{20}P- 2000*\displaystyle\sum_{k=0}^{19}\;1.04^{19-k}*1.03^k\\ 2000*\displaystyle\sum_{k=0}^{19}\;1.04^{19-k}*1.03^k = 1.04^{20}P\\ 1.04^{20}P=2000*\displaystyle\sum_{k=0}^{19}\;1.04^{19-k}*1.03^k \\ P=\left[2000*\displaystyle\sum_{k=0}^{19}\;1.04^{19-k}*1.03^k \right]\div[ 1.04^{20}]\\\)

Initial value was $35,142.88          

Find the 2015th term of the geometric sequence 1,−1/3,-1/6,−1/9 ,.

In a geometric sequence, the common ratio, r between any two consecutive terms is always the same.

\(\begin{array}{rcl} \dfrac{ \text{Any term} } { \text{Previous term}} &=& \dfrac{u_n}{ u_n-1} = \text{ constant } = r \\ \end{array}\)

\(\begin{array}{rcr} 1,-\frac13,-\frac16,-\frac19,-\cdots \\ u_1 &=& 1\\ u_2 &=& -\frac13\\ u_3 &=& -\frac16\\ u_4 &=& -\frac19\\ \cdots \\ \end{array} \)

If three terms,  \(u_n,\ u_{n+1},\ u_{n+2}\) are in geometric sequence, then:

\(\begin{array}{rcl} \dfrac{ u_{n+2} } { u_{n+1} } &=& \dfrac{ u_{n+1} } { u_{n} }\\ \end{array}\)

We have

\(\begin{array}{rcr} \dfrac{ u_2 } { u_1 } = \dfrac{ -\frac13 } { 1 } = -\frac13 &\ne& \dfrac{ u_3 } { u_2 } = \dfrac{ -\frac16 } { -\frac13 } = \frac12 \\ &\ne& \dfrac{ u_4 } { u_3 } = \dfrac{ -\frac19 } { -\frac16 } = \frac23 \\ \end{array}\)

Sorry, this is not a geometric sequence.

I can't DRAW the Quadratic Formula, but it is

(   -b +- sqrt( b^2 -4ac) )  /  2a

In this equation   a = 1   b=11  c= -48   

Substituting:

 (-11 +- sqrt(121 - (-192)) / 2  =  -11 +- sqrt313 =  (-11 +-17.69) / 2 = -28.69/2    and    6.69/2       or   -14.35    and 3.35

As a check, you can substitute these values BACK into the ORIGINAL equation to see if they are true....

Fri 27/11/15

1) Call for guests to become members.   Thanks Anonymous4338  :)

https://web2.0calc.com/questions/web-2-0-calc

2) How do you know if a number is prime?  Thanks Gino3141 and guest.  :)

https://web2.0calc.com/questions/prime-numbers_2

3) This was in yesterday and I would still like someone to discuss it.  :/

https://web2.0calc.com/questions/calculus-question_1

4) Inequality    Thanks Anonymous43338

https://web2.0calc.com/questions/mymaths

5) Function of a function.     Thanks Geno and Heureka

https://web2.0calc.com/questions/f-f-x-x-type-of-question#r2

6) Proof      Thanks Heureka

https://web2.0calc.com/questions/if-n-is-a-positive-integer-such-that-2n-1-is-a-perfect-square-show-that-n-1-is-the-sum-of-two-successive-perfect-squares#r1

7) Proof    Thanks guys but I would really like another mathematician to look at this one. :/

https://web2.0calc.com/questions/help_67408

I'm pretty sure you'll need to tell us:  For how long?

@@ What is Happening?  [Wrap4]   Thurs 26/11/15   Sydney, Australia Time 9:38 pm   ♪ ♫

Hi all,

Our great answers have been supplied by CPhill, Heureka, Alan, Riddler, gibsonj338, Olivia3.141592654, Buubleman, Bertie, Anonymous4338, Steven434, Hayley, jc and guests.    Thank you.   

Interest Posts:

If you ask or answer an interesting question, you can private message the address to me (with copy and paste) and I will include it.  Of course only members are able to do this.  I quite likely will not see it if you do not show me.  

1)  Fun facts about 666     Thanks CPhill

     One has to wonder about Chris's motives LOL

1.) Angle X and Angle Y are supplementary. Find the measure of each angle if Angle Measurement X= 6x -1 and Angle MEasurement Y=5x -17.

Two angles that sum to a straight angle \(180^{\circ} \)are called supplementary angles.

\(\begin{array}{rcl} X &=& 6x-1 \\ Y &=& 5x-17 \\ X+Y &=& 180^{\circ} = 6x-1 + 5x-17\\ 180^{\circ} &=& 11x-18\\ 180^{\circ}+18 &=& 11x\\ 198 &=& 11x\\ 11x &=& 198 \\ x &=& \frac{198}{11} \\ x &=& 18 \\\\ X &=& 6x -1\\ X &=& 6\cdot 18 -1 \\ X &=& 107^{\circ}\\\\ Y &=& 180^{\circ} -X \\ Y &=& 180^{\circ} -107^{\circ}\\ Y &=& 73^{\circ} \end{array}\)

2.) /P and /Q are complementary. The measure of /Q is 4 times the measure of /P. Find the measure of each angle.

Complementary angles are angle pairs whose measures sum to one right angle \( 90^{\circ}.\)

\(\begin{array}{rcl} Q &=& 4P\\ P + Q &=& 90^{\circ}\\ P + 4P &=& 90^{\circ}\\ 5P &=& 90^{\circ}\\ P &=& \frac{ 90^{\circ} }{5}\\ P &=& 18^{\circ}\\\\ Q &=& 90^{\circ} - P \\ Q &=& 90^{\circ} - 18^{\circ}\\ Q &=& 72^{\circ}\\ \end{array}\)

Thank you Kitty <3         

What is the answer to square root 10 over 2 doubled?

\(2\times \frac{\sqrt{10}}{2}=\sqrt{10}\)

Very very Nice proof Berti !

Your answers are always wonderful

\((4+32i)^(1/5)\)

\(r=\sqrt(a^2+b^2)\)

\(r=\sqrt(4^2+32^2)\)

\(r=\sqrt(16+1024)\)

\(r=\sqrt1040\)

\(r=4\sqrt65\)

\(tan(\Theta)=b/a\)

\(tan(\Theta)=32/4\)

\(tan(\Theta)=8\)

\(\Theta=tan^-1(8)\)

\(\Theta ≈1.4464413322481\)

\(z=r*e^(i*\Theta)\)

\(z≈4\sqrt65*e^(i*1.4464413322481)\)

\(z^(1/5)≈(4\sqrt65*e^(i*1.4464413322481))^(1/5)\)

\(z^(1/5)≈(4\sqrt65)^(1/5)*e^(i*1.4464413322481*(1/5))\)

\(z^(1/5)≈2.003103242348*e^(i*1.0766143512748)\)

\(z=r*(cos(\Theta)+i*sin(\Theta))\)

\(z≈2.003103242348*(cos(1.0766143512748)+i*sin(1.0766143512748))\)

\(z≈2.003103242348*(0.4743116564868+i*0.88035700289064)\)

\(z≈0.9500952169545+i*1.763445966914\)

\(z≈0.9500952169545+1.763445966914i\)

\(z^(1/5)≈(4\sqrt65*e^(i*7.7296266394277))^(1/5)\)

\(z^(1/5)≈(4\sqrt65)^(1/5)*e^(i*7.7296266394277*(1/5))\)

\(z^(1/5)≈2.003103242348*e^(i*1.5459253278855)\)

\(z=r*(cos(\Theta)+i*sin(\Theta))\)

\(z≈2.003103242348*(cos(1.5459253278855)+i*sin(1.5459253278855))\)

\(z≈2.003103242348*(0.024868434927217+i*0.99969073264899)\)

\(z≈-0.049814942634829+i*0.0001545372385216\)

\(z≈-0.049814942634829+0.0001545372385216i\)

\(z^(1/5)≈(4\sqrt65*e^(i*14.012811946607))^(1/5)\)

\(z^(1/5)≈(4\sqrt65)^(1/5)*e^(i*14.012811946607*(1/5))\)

\(z^(1/5)≈2.003103242348*e^(i*1.6955283616309)\)

\(z=r*(cos(\Theta)+i*sin(\Theta))\)

\(z≈2.003103242348*(cos(1.6955283616309)+i*sin(1.6955283616309))\)

\(z≈2.003103242348*(-0.12440885450163+i*0.99223104009177)\)

\(z≈-0.24920377983349+i*1.9875412136019\)

\(z≈-0.24920377983349+1.9875412136019i\)

\(z^(1/5)≈(4\sqrt65*e^(i*20.2959972538))^(1/5)\)

\(z^(1/5)≈(4\sqrt65)^(1/5)*e^(i*20.2959972538*(1/5))\)

\(z^(1/5)≈2.003103242348*e^(i*4.059994507574)\)

\(z=r*(cos(\Theta)+i*sin(\Theta))\)

\(z≈2.003103242348*(cos(4.05994507574)+i*sin(4.05994507574))\)

\(z≈2.003103242348*(-0.60772245598411+i*-0.7941494925344)\)

\(z≈-1.2173308220295+i*1.5907634234047\)

\(z≈-1.2173308220295+1.5907634234047i\)

\(z^(1/5)≈(4\sqrt65*e^(i*26.5918256098))^(1/5)\)

\(z^(1/5)≈(4\sqrt65)^(1/5)*e^(i*26.5918256098*(1/5))\)

\(z^(1/5)≈2.003103242348*e^(i*5.315836512196)\)

\(z=r*(cos(\Theta)+i*sin(\Theta))\)

\(z≈2.003103242348*(cos(5.315836512196)+i*sin(5.315836512196))\)

\(z≈2.003103242348*(0.56748448302716+i*-0.82338409112843)\)

\(z≈-1.1367300079339+i*-1.6493233426371\)

\(z≈-1.1367300079339+-1.6493233426371i\)

\(z≈-1.1367300079339-1.6493233426371i\)

32 1/2 divided by 1/5...

First, convert the mixed number, 32 1/2, to an improper fraction (multiply the whole number and the denominator, then add that to the numerator. In this case, 32 1/2 becomes 65/2. This gets you an equation of 65/2 / 1/5.

Next, use "keep, change, flip."

Keep 65/2.

Change the division sign into a multiplication sign.

Flip 1/5 to 5/1.

You now have 65/2*5/1.

From here, just multiply the numerators together to get your new numerator, and the same but for your denominators.

65*5=325

2*1=2

This gets you 325/2. Simplify, and you get 162.5.

In conclusion, 32.5 divided by 1/5 is equal to 162.5, or 162 1/2.

It's way past your bed time, Anon4338

Okay, let's divide this problem up.

2x+5>10

Subtract 5 from both sides...

2x>5

Divide by 2 on each side...

x>2.5

This means that we know that x is greater than 2.5. Next, let's do 5x-4<20.

5x-4<20

Add 4 to each side...

5x<24

Divide by 5 on each side...

x<4.8

This means that x is less than 4.8.

Now we know that x is greater than 2.5, but less than 4.8.

2.5<x<4.8.

In conclusion, x can be any number from 2.5 to 4.8.

Good job, Guest!

63 days divided by 7 days a week equals 9 weeks.

True, but whenever you are adding two negatives together, you can just subtract the absolute value of the second number from the first.

-10+(-30) turns into -10-30. The answer is -40, as Steven434 said. You can access the calculator on this website by clicking on the home tab at the top of the page.

If you mean 1/7(6207/5400), then you can just multiply the numerators to get your new numerator, and  multiply the denominators to get your new denominators. 1*6207=6207, and 7*5400=37800. Your answer is 6207/37800.

An investment allows for end-of-year payments that grow by 3% per payment for 20 years, starting with $2000 at the end of the first year. If the payments earn 4% compounded annually, how much was invested at the beginning

This is the the most complicated of all the problems seen so far. There are at least two different formulae used in this problem. They are relatively complicated, because you have to take into account the increase of 3% in the regular annual payment of $2,000 per year. It is essentially the same as taking inflation into account, in that your payment increases by 3% every year. Based on all that the PV of the initial investment comes to=$35,142.88. By the way, for your teacher to give you these problems, the formulae must in your book, properly explained. You may have to look them up.

I have the advantage of having financial software on my computer that can solve all these problems almost instantaneously.

I don't really understand your question