okay thank you hectictar

Deleted.

I do not get that either Alan .  

Hi all! i a having trouble figurig out the math of this..

I have 30.000 on my savings account. It has an interest rate of 0.007 percent. Every month for a year I add 5000 to my savings account.

How much money will I have by the end of the year?

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How often is the interest added?  If it is added monthly (which is never really the case) then the interest rate is  probably  0.007/12 % per month

Plus I doubt that you mean 0.007%, I expect you mean 0.007 which is 0.7%

So let me assume that you get  0.7/12 % monthly.  = 0.000583 repeater

I will also  assume that the the $5000 is deposited at the END of every month, including the last month.

\(30000*(1+\frac{0.0007}{12})^{12}+5000[(1+\frac{0.0007}{12})^{11}+(1+\frac{0.0007}{12})^{10}+...(1+\frac{0.0007}{12})^{0}]\\ 30000*(1.0005833333)^{12}+5000[(1.0005833333)^{11}+(1.0005833333)^{10}+...(1.0005833333)^{0}]\\ 30000*(1.0005833333)^{12}+5000[(1.0005833333)^{11}+(1.0005833333)^{10}+...(1.0005833333)^{0}]\\\)

You can work out the bit in the squar brackets using the sum of a GP formula.

There are formula to do this but this is how they are derived.

How did you get that equation that would give you the largest value of a?

Also, why can't OB = 2, for example?

If a, b, and c are positive integers less than 13 such that
\(\large{\begin{align*} 2ab+bc+ca&\equiv 0\pmod{13}\\ ab+2bc+ca&\equiv 6abc\pmod{13}\\ ab+bc+2ca&\equiv 8abc\pmod {13} \end{align*}}\)
then determine the remainder when a+b+c is divided by 13.

\(\begin{array}{|lrcll|} \hline & 2ab+bc+ca &\equiv& 0\pmod{13} \quad |\quad : (abc)\\ (1) & \dfrac{2}{c} + \dfrac{1}{a} + \dfrac{1}{b} &\equiv& 0 \pmod{13} \\ \hline & ab +2bc+ ca &\equiv& 6abc\pmod{13} \quad |\quad : (abc)\\ (2) & \dfrac{1}{c} + \dfrac{2}{a} + \dfrac{1}{b} &\equiv& 6 \pmod{13} \\ \hline & ab + bc+2ca &\equiv& 8abc\pmod{13} \quad |\quad : (abc)\\ (3) & \dfrac{1}{c} + \dfrac{1}{a} + \dfrac{2}{b} &\equiv& 8 \pmod{13} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline 3*(1)-(2)-(3): & 3*\left( \dfrac{2}{c} + \dfrac{1}{a} + \dfrac{1}{b}\right) \\ & -\left( \dfrac{1}{c} + \dfrac{2}{a} + \dfrac{1}{b}\right) \\ & -\left( \dfrac{1}{c} + \dfrac{1}{a} + \dfrac{2}{b}\right) \\ & &\equiv& 3*0 - 6 - 8 \pmod{13} \\ & \dfrac{6}{c} + \dfrac{3}{a} + \dfrac{3}{b} \\ & -\dfrac{1}{c} - \dfrac{2}{a} - \dfrac{1}{b} \\ & -\dfrac{1}{c} - \dfrac{1}{a} - \dfrac{2}{b} \\ & &\equiv& -14 \pmod{13} \\ & \dfrac{4}{c} &\equiv& -14 \pmod{13} \\ & \dfrac{4}{c} &\equiv& 13-14 \pmod{13} \\ & \dfrac{4}{c} &\equiv& -1 \pmod{13} \quad | \quad +1 \\ & \dfrac{4}{c}+1 &\equiv& 0 \pmod{13} \quad | \quad *c \\ & 4+c &\equiv& 0 \pmod{13} \quad | \quad -4 \\ & c &\equiv& -4 \pmod{13} \\ & c &\equiv& 13-4 \pmod{13} \\ & \mathbf{ c } &\equiv& \mathbf{ 9 \pmod{13} } \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline 3*(3)-(2)-(1): & 3*\left( \dfrac{1}{c} + \dfrac{1}{a} + \dfrac{2}{b}\right) \\ & -\left( \dfrac{1}{c} + \dfrac{2}{a} + \dfrac{1}{b}\right) \\ & -\left( \dfrac{2}{c} + \dfrac{1}{a} + \dfrac{1}{b}\right) \\ & &\equiv& 3*8 - 6 - 0 \pmod{13} \\ & \dfrac{3}{c} + \dfrac{3}{a} + \dfrac{6}{b} \\ & -\dfrac{1}{c} - \dfrac{2}{a} - \dfrac{1}{b} \\ & -\dfrac{2}{c} - \dfrac{1}{a} - \dfrac{1}{b} \\ & &\equiv& 18 \pmod{13} \\ & \dfrac{4}{b} &\equiv& 18-13 \pmod{13} \\ & \dfrac{4}{b} &\equiv& 5 \pmod{13} \quad | \quad -5 \\ & \dfrac{4}{b}-5 &\equiv& 0 \pmod{13} \quad | \quad *b \\ & 4-5b &\equiv& 0 \pmod{13} \quad | \quad *(-1) \\ & 5b &\equiv& 0 \pmod{13} \quad | \quad +4 \\ & 5b &\equiv& 4 \pmod{13} \quad | \quad : 5 \\ & b &\equiv& 4*\dfrac{1}{5} \pmod{13} \\ & && \boxed{\dfrac{1}{5} \pmod{13} \\ \equiv 5^{\varphi(13)-1}\pmod{13} \\ \equiv 5^{12-1}\pmod{13} \\ \equiv 5^{11}\pmod{13} \\ \equiv 48828125\pmod{13} \\ \equiv 8\pmod{13} } \\ & b &\equiv& 4*8 \pmod{13} \\ & b &\equiv& 32 \pmod{13} \\ & b &\equiv& 32-2*13 \pmod{13} \\ & b &\equiv& 6 \pmod{13} \\ & \mathbf{ b } &\equiv& \mathbf{ 6 \pmod{13} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 2ab+bc+ca &\equiv& 0\pmod{13} \quad & | \quad b=6, \ c=9 \\ 2a*6+6*9+9*a &\equiv& 0\pmod{13} \\ 21a+54 &\equiv& 0\pmod{13} \quad & | \quad 21\equiv 8\pmod{13},\ \quad 54\equiv 2\pmod{13} \\ 8a +2 &\equiv& 0\pmod{13} \quad & | \quad :2 \\ 4a+1 &\equiv& 0 \pmod{13} \quad & | \quad -1 \\ 4a &\equiv& -1 \pmod{13} \quad | \quad : 4 \\ a &\equiv& (-1)*\dfrac{1}{4} \pmod{13} \\ && \boxed{\dfrac{1}{4} \pmod{13} \\ \equiv 4^{\varphi(13)-1}\pmod{13} \\ \equiv 4^{12-1}\pmod{13} \\ \equiv 4^{11}\pmod{13} \\ \equiv 4194304\pmod{13} \\ \equiv 10\pmod{13} } \\ a &\equiv& (-1)*10 \pmod{13} \\ a &\equiv& -10 \pmod{13} \\ a &\equiv& 13-10 \pmod{13} \\ \mathbf{ a } &\equiv& \mathbf{ 3 \pmod{13} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{a+b+c \pmod{13}} \\ &\equiv& 3+6+9 \pmod{13} \\ &\equiv& 18 \pmod{13} \\ &\equiv& 18-13 \pmod{13} \\ &\equiv& \mathbf{ 5 \pmod{13} } \\ \hline \end{array}\)

Here's a rough and ready way to tackle this:

Notice that if b = 2a and c = 3a the left hand side of the first equation becomes 4a² + 6a² + 3a² = 13a²

In other words it is a multiple of 13 and hence satisfies the first equation.

For this situation, if a, b and c are all positive integers less than 13, then a can only be one of 1, 2, 3 or 4.

Trying these in turn we find that only a = 3 (hence b = 6 and c = 9) satisfy the second and third equations.

Hence a + b + c  =  5 mod(13)  

As follows:

Let side OA (opposite angle B) be b, and side OB (opposite angle A) be a.

From the cosine rule we have:  \(1^2=a^2+b^2-2ab\frac{\sqrt3}{2}\)  remembering that \(cos(30\deg)=\frac{\sqrt3}{2}\).

Rewrite as a quadratic in a:  \(a^2-\sqrt3ba+b^2-1=0\)

Solve for a in terms of b:  \(a=\frac{\sqrt3}{2}b+\frac{\sqrt{(2+b)(2-b)}}{2}\)  (This solution will give the largest value of a)

Differentiate a wrt b to get:  \(\frac{da}{db}=\frac{\sqrt3}{2}-\frac{b}{2\sqrt{(2+b)(2-b)}}\)

Set this to zero and solve for b to get:  \(b=\sqrt3\)

Substitute this back into the expression for a above to get:  \(a=2\)

(Corrected for earlier brain fade!!!).

We can find the length of the missing sides using the information that the area is 240 sq units. 

area of triangle  =  ( 1/2 )( base )( height )     Plug in the information we're given.

240  =  ( 1/2 )( 20 )( height )     Multiply  1/2   by   20

240  =  ( 10 )( height )          Divide both sides of the equation by  10

24  =  height

So we know the base is  20  and the height is  24

When we draw a height to the base, we split the triangle into two smaller congruent triangles.

(We can know they're congruent by the hypotenuse-leg congruence theorem.)

And so the height splits the base exactly in half.

That means the length of the missing leg in each of the right triangles  =  20/2  =  10

So by the Pythagorean Theorem,

10² + 24²  =  c²

100 + 576  =  c²

676  =  c²

c  =  √676

c  =  26

Now do you see how to finish it?

{ a^-1 }  =  { a² }

Using the definition of the  { }  function, we can say

a^-1 - floor( a^-1 )  =  a² - floor( a² )

Now since  2 < a² < 3 ,  a  can't be  1 .  And so  floor( a^-1 )  =  0

Also since  2 < a² < 3 ,  we can say for sure that  floor( a² )  =  2

a^-1 - 0  =  a² - 2

                              Subtract  a^-1  from both sides of the equation.

0  =  a² - 2 - a^-1

                              Multiply through by  a  and note  a ≠ 0

0  =  a³ - 2a - 1

0  =  a³ - a  -  a - 1

0  =  a³ - a² - a  +  a² - a - 1

0  =  a(a² - a - 1) + 1(a² - a - 1)

0  =  (a + 1)(a² - a - 1)

We only want the middle solution because the other two violate the inequality  2 < a² < 3

So     a  =  (1 + √5 ) / 2

thanks!

An inscribed angle is half the measure of its intercepted arc, so

m∠ADC  =  (1/2)(m∠ABC)

                                              Substitute  23°  in for  m∠ADC

23°  =  (1/2)(m∠ABC)

                                              Multiply both sides by  2

46°  =  m∠ABC

There are 3 pairs that satisfy the conditions given in the question and are as follows:
G =5   and H =1
G =1   and H =5
G =6    and H=9
So, the distinct product values are:
5 x 1 = 5
6 x 9 = 54
Sum of the 2 distinct products =5 + 54 = 59 

There's no fancy way to repost a question. Just make a new question and put your old question there, but please make sure to put a link to the original question too. And it would be a nice idea to add a sentence explaining why you're reposting it.

To post an answer to an old question that has been locked, you'll have to ask a moderator to unlock it for you.

By the SAS congruence theorem,  △PQT ≅ △SRT  So...

m∠QPT  =  m∠RST  =  37°

m∠QTP  =  180° - 50°   =   130°

m∠PQT  =  180° - 37° - 130°  =  13°

Since  QT = RT  we can substitute  QT  in for  RT in the next equation.

\(RT+TP\ =\ 11\\~\\ QT+TP\ =\ 11\\~\\ QT\ =\ 11 - TP\)

Now we can substitute  11 - TP  in for  QT in the next equation.

By the Law of Sines,

\( \frac{TP}{\sin13^\circ} \ =\ \frac{QT}{\sin37^\circ}\\~\\ TP\cdot\frac{\sin37^\circ}{\sin13^\circ}\ =\ QT\\~\\ TP\cdot\frac{\sin37^\circ}{\sin13^\circ}\ =\ 11-TP\\~\\ TP\cdot\frac{\sin37^\circ}{\sin13^\circ}+TP\ =\ 11\\~\\ TP\cdot(\frac{\sin37^\circ}{\sin13^\circ}+1)\ =\ 11\\~\\ TP\ =\ 11\div(\frac{\sin37^\circ}{\sin13^\circ}+1)\\~\\ TP\ =\ \frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ} \)

Now that we know the length of  TP ,  we can find the length of  QT.

\(QT\ =\ 11-TP\\~\\ QT\ =\ 11-\frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}\)

Finally, we can use the Law of Sines again to find  QP.

\(\frac{QP}{\sin130^\circ}\ =\ \frac{TP}{\sin13^\circ}\\~\\ QP\ =\ TP\cdot\frac{\sin130^\circ}{\sin13^\circ}\\~\\ QP\ =\ \frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}\cdot\frac{\sin130^\circ}{\sin13^\circ}\\~\\ QP\ =\ \frac{11\sin130^\circ}{\sin37^\circ+\sin13^\circ}\)

So we have found:

\(\begin{array}{c} TP&=& \frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}& \approx& 2.993\\~\\ QT& =& 11-\frac{11\sin13^\circ}{\sin37^\circ+\sin13^\circ}& \approx& 8.007\\~\\ QP& =&\frac{11\sin130^\circ}{\sin37^\circ+\sin13^\circ}& \approx& 10.192 \end{array}\)

And all lengths are in meters.

\(here are $13$ people in Herm's group. The only purpose of the detailed explanation of the ``handshake'' is to illustrate that this is not a normal handshake, and that there is a distinct role for each of the three people involved in the handshake. There are $13$ possibilities for the hook, $12$ possibilities for the bait, and $11$ possibilities for the bandit. Therefore, we have $13 \cdot 12 \cdot 11 = \boxed{1716}$ total possibilites.\)

to upload a pic you have to click on the IMAGE button and insert the url of the image or you can up load an image off the computer, phone, or other divices.

😉

Travisio

Domain = all real numbers

Range = all real numbers

I am not sure of waht you are asking, but:

5% of 320 -= .05 (320) = .05*320 = 16

or 

320 is 5% of  ?

.05x= 320

x = 320/.05 = 6400

Well you will have tje same answer as us then anyway.    

I was taught, and I believe it is proper, that when the decimal fraction is exactly .5 then if the number preceding it is odd it rounds up, and if the number preceding it is even it rounds down. 

Good for you. You were taught correctly.  The rounding method you describe is still used in statistical analysis. 

The floor function for odd numbers and the ceiling function for even numbers that are followed by an exact value of (5) are used to process collected data to prevent rounding bias for statistical analysis, and other data processing reasons. One notable exception is the data used for interest payouts for financial instruments, such as dividends on stocks, bank accounts, etc.  An ending “5” will always use the ceiling function regardless if the preceding number is even or odd. The reasons for this exception are interesting. 

This however is a type of Diophantine question –meaning that it requires integers for solutions. Though derived from a statistical perspective, the nature of the question presented is minimally statistical, and the data used to solve it is not statistical –the variables are not random. The writer of the question included a parenthetical statement to clarify the exception for this question.

GA

I thought it strange that all these angles should be vertical. Logically, half should be “horizontal” angles.

Below is an image excerpt from Methods for Euclidean Geometry –37^th Ed.  (Copyright 2010)

The text above the theorem [seems to] indicate[s] all the angles are vertical.

The theorem itself indicates only two are vertical.  It doesn’t indentify the other angles, only that they are also congruent. 

Next time, I’ll divide by 2. 

GA

If you want to know how much you'd have to add to  340⁴  to get  680⁴, we can solve this equation for  x:

x + 340⁴  =  680⁴

x  =  680⁴ - 340⁴

x  =  200450400000

So

200450400000 + 340⁴  =  680⁴

In other words,  680⁴  is  200450400000  more than  340⁴

If you want to know what power must  340⁴  be raised to to equal  680⁴,  we can solve this equation for  x:

(340⁴)^x   =   680⁴

340^x  =  680

x  ≈  1.1189

So

(340⁴)^1.1189  ≈  680⁴

680⁴   =   (2 * 340)⁴

680⁴   =   2⁴ * 340⁴

680⁴   =   16 * 340⁴

In other words,  680⁴  is  16  times  340⁴