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How many 10 digit numbers are there, such that the sum of the digits is divisible by 5?

The first digit cannot be zero, so 9 ways, the next 8 digits can be anything, so there are 10 ways for each of them.
We can then only choose two digits as last digit to satisfy the condition.

There are \(\mathbf{9\times 10^8 \times 2 = 1~ 800~ 000~ 000}\) numbers.

\(\begin{array}{|c|c|} \hline \text{the sum of the digits 1 until 9} & \text{last digit to satisfy the condition }\\ \hline \ldots 0 & 0~ \text{ or } ~5 \\ \ldots 1 & 4~ \text{ or } ~9 \\ \ldots 2 & 3~ \text{ or } ~8 \\ \ldots 3 & 2~ \text{ or } ~7 \\ \ldots 4 & 1~ \text{ or } ~6 \\ \ldots 5 & 0~ \text{ or } ~5 \\ \ldots 6 & 4~ \text{ or } ~9 \\ \ldots 7 & 3~ \text{ or } ~8 \\ \ldots 8 & 2~ \text{ or } ~7 \\ \ldots 9 & 1~ \text{ or } ~6 \\ \hline \end{array}\)

Example: 1 000 000 08(1) or 1 000 000 08(6)

               4 567 123 98(0) or 4 567 123 98(5)

The 1 big one, with nine little squares inside it

The 1 at the top left corner, with four little squares inside it

The 3 other corners, each with four little squares inside them

The 9 little squares

Total 14   

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that is incorrect

that is incorrect

that is incorrect

What is the perimeter?

3/4 of $600 = $450 

(1-0.2=0.8)

80% of $450 = $360 (the lemons she sold at a loss)

1/4 of 600 = 150

(1+0.2=1.2)

120% of 150=$180 (the lemons she sold at a gain)

180+360=540

540/600= 90% 

100-90= 10% 

She lost 10% in the whole transaction.

I hope this is right or this will be awkward :)

It is correct.

Omi67

C=6R

C+2=4(R+2)

now solve those equations simultaneously.

This was a good geometry question with some equally interesting answers.

What are the coordinates of the foci of the hyperbola?

The general form of a parabola is y^2-4x+4y-8=0

What is the standard form?

Thanks guest for the complement as well as for the explanation.

I am all clear now.

Thanks Heureka and Alan as well      

How many squares can you find in this picture ?

Hello Guest, I see 14 squares.

  !

The foci are at (-8,-2) and (-8,8).

Three runners Adam, Ben, and Charles all start at the same time for a 24 km race, and each of them runs at a constant speed. When Adam finishes the race, Ben is 8 km behind, and Charles is 12 km behind. When Ben finishes the race, how far behind is Charles, in km?

Drei Läufer, Adam, Ben und Charles, starten alle gleichzeitig für ein 24-km-Rennen, und jeder von ihnen läuft mit einer konstanten Geschwindigkeit. Als Adam das Rennen beendet hat, liegt Ben 8 km und Charles 12 km zurück. Wenn Ben das Rennen beendet, wie weit ist Charles zurück.

Hello Guest!

\(v=\frac{s}{t}\\ t=\frac{s}{v}\\ s=vt\)

\(\frac{24}{v_a}=\frac{16}{v_b}=\frac{12}{v_c}\)

\(\frac{v_b}{16}=\frac{v_c}{12}\\ v_c=\frac{3}{4}v_b\)

\(\frac{24km}{v_b}=\frac{24km-x}{v_c}\\ \frac{24km}{v_b}=\frac{24km-x}{\frac{3}{4}v_b}\\ 3\cdot 24km=4(24km-x)\\ 72km=96km-4x\\ 4x=(96-72)km\)

\(x=6km\)

Charles is 6km behind.

  !

OK... lets say adam runs at 24 km /hr  finishes in 1 hour

              charles is 12 km behind .....he is only 1/2 way through....must be running  at 12 km/hr   he will finish in 2 hours

                  ben is 8km behind    he has run 16 km in one hour     16 km/hr    he will finish in 1  1/2 hour

                        Charles  done in 2 hours

                               whne Ben finishes Charles is   still  30 minutes behind x 12/km/hr  = 6 km

Three runners Adam, Ben, and Charles all start
at the same time for a 24 km race,
and each of them runs at a constant speed.
When Adam finishes the race, Ben is 8 km behind, and Charles is 12 km behind.
When Ben finishes the race, how far behind is Charles, in km?

Formula \(s=v*t\)

\(\begin{array}{|lrcll|} \hline (1) & 24~\text{km} &=& v_{Adam}*t_{Adam} \\ (2) & 24~\text{km}-8~\text{km} &=& v_{Ben}*t_{Adam} \\ &\mathbf{ v_{Ben}} &=& \mathbf{\dfrac{16}{t_{Adam}} } \\\\ (3) & 24~\text{km}-12~\text{km} &=& v_{Charles}*t_{Adam} \\ & \mathbf{v_{Charles}} &=& \mathbf{ \dfrac{12}{t_{Adam}} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline s_{Ben}=24~\text{km} &=& v_{Ben}*t_{Ben} \quad |\quad \mathbf{ v_{Ben}=\dfrac{16}{t_{Adam}} } \\ \\ 24 &=& \dfrac{16}{t_{Adam}}*t_{Ben} \\\\ \dfrac{t_{Ben}}{t_{Adam}} &=& \dfrac{24}{16} \\ \\ \mathbf{ \dfrac{t_{Ben}}{t_{Adam}} } &=& \mathbf{1.5 } \\\\ \hline s_{Charles} &=& v_{Charles}*t_{Ben} \quad |\quad \mathbf{v_{Charles}=\dfrac{12}{t_{Adam}} } \\ \\ s_{Charles} &=& \dfrac{12}{t_{Adam}}*t_{Ben} \\ \\ s_{Charles} &=& 12*\dfrac{t_{Ben} }{t_{Adam}} \quad | \quad \mathbf{ \dfrac{t_{Ben}}{t_{Adam}} = 1.5 } \\\\ s_{Charles} &=& 12* 1.5 \\\\ s_{Charles} &=& 18~\text{km} \\ \hline s_{Ben}-s_{Charles} &=& 24~\text{km}-18~\text{km}\\ \mathbf{s_{Ben}-s_{Charles}} &=& \mathbf{6~\text{km}} \\ \hline \end{array}\)

Charles is \(\mathbf{6~\text{km}}\) behind

6 gallons / 12 hours =      1/2 gallon/hour

1/2 gallon/hour x 42 hours = 21 gallons lost 

  $  600 x 3/4 = 450  at a loss 0f 20%  so she received;   450 x .8 = $ 360

       600-450 = $ 150 lemons left   

  $  150  x 1.20   = $ 180

180 + 360 = 540     she lost  60$ overall     60/600 = 10%

EDITED TO CORRECT MISTAKE !!!  1/19/20

At a party. Andy and April shared a cherry pie. Andy ate 2/5 of the pie and April ate 1/3 of the pie.

What fraction of the pie did they eat altogether

\(\begin{array}{|rcll|} \hline &&\mathbf{ \dfrac{2}{5} + \dfrac{1}{3} } \\\\ &=& \dfrac{2*3+1*5}{5*3}\\\\ &=& \mathbf{ \dfrac{11}{15}} \\ \hline \end{array}\)

A coffee shop currently sells 490 lattes a day at $3.00 each.

They recently tried raising the by price by $0.25 a latte, and found that they sold 30 less lattes a day.

a) Assume that the number of lattes they sell in a day, N, is linearly related to the sale price, p (in dollars). Find an equation for N as a function of p.

\(\begin{array}{|rcll|} \hline \dfrac{N-460}{p-3.25} &=& \dfrac{490-460}{3-3.25} \\\\ N-460 &=& (p-3.25)*\dfrac{490-460}{3-3.25} \\\\ N &=& 460+(p-3.25)*\dfrac{490-460}{3-3.25} \\\\ N &=& 460+(p-3.25)*\dfrac{30}{-0.25} \\\\ N &=& 460-(p-3.25)*\dfrac{30}{0.25} \\\\ N &=& 460-p*\dfrac{30}{0.25}+3.25*\dfrac{30}{0.25} \\\\ N &=& 460-p*\dfrac{30}{0.25}+390 \\\\ N &=& 850-p*\dfrac{30}{0.25} \\\\ N &=& 850-p*120 \\\\ \mathbf{N} &=& \mathbf{850 - 120p} \\ \hline \end{array}\)

If the farm has 30 chickens and cows,
and there are 82 chicken and cow legs all together,
then how many chickens and how many cows could the farm have

\(\begin{array}{|rcll|} \hline ch + co &=& 30 \\ ch &=& 30-co \\\\ 2*ch + 4*co &=& 82 \\ 2*(30-co) + 4*co &=& 82 \\ 60-2*co + 4*co &=& 82 \\ 60 + 2*co &=& 82 \\ 2*co &=& 82-60 \\ 2*co &=& 22 \quad | \quad :2 \\ \mathbf{\text{cows}} &=& \mathbf{11} \\\\ ch &=& 30-co \\ ch &=& 30-11 \\ \mathbf{\text{chickens}} &=& \mathbf{19} \\ \hline \end{array}\)

Call the amount that Mahmud has              X      

And call the amount that Rihan has is        4 • 1/5 • X    

Together, they have                                    756

Add M's & R's together to equal                  X + ( 4 • 1/5 • X )  =  756

4 • 1/5 • X = 4X/5

1 • X          = 5X/5

                                                                    4X/5 + 5X/5  =  9X/5  =  756      

Multiply both sides by 5                                                           9X  =  3,780     

Divide both sides by 9                                                               X  =  420        

Mahmoud has 420 

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1. R and S have 56 elements in common.

4. The expression is equal to -4.