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10 ca we changed into 8 different ways:-

10

5,5

5,2,2,1

2,2,2,2,1,1

2,2,2,1,1,1,1

2,2,1,1,1,1,1,1

2,1,1,1,1,1,1,1,1

8 1s

Since 10,10 counts as 1

so we know that there will be a total of 8+7 different ways to change

so your answer is 15

x = 1.5 y     or    y = 2/3 x

x = .6z        or    z = 1 2/3 x

x                                                       summed, the red equations = 80

2/3 x + 1 2/3 x   + x = 80

3 1/3 x = 80

x = 24 sudents

2^10=1024

2+4+8+16+32+64+128+256+512=1022

1024-1022=2

Each ball has five options to go 

Therefore four balls  will be filled in =

5×5×5×5

= 5^4 ways = 625 ways

Each ball has five options to go 

Therefore four balls  will be filled in =

5×5×5×5

= 5^4 ways = 625 ways

x=3y/2 x=3z/5 z=15y/6=5y/2

y+3y/2+5y/2=10y/2=5y y=16

x=24 z=40

40+24+16=80

x=24

Thank you!

Key idea:

1) If the given sequence does have common difference, or common ratio, then it is neither an arithmatic nor geometric.

2) How to find out :

We say "d" is common difference of arithmatic :

If d value is   (a2 - a1) ≠ (a3 - a2 )  then it is not arithmatic sequences.

We say "r" is common ratio of geomatric sequences:

If r value is  (a₁ / a ₂) ≠ (a₂ / a₃ ) , then it is not geomatric sequences. 

...  Here, we get common ratio of geomatric sequence =  (1/4) /(1/6) = (1/6) /(1/9) = (3/2)  , Then it is geometric sequences.

3) Is  (3/2) greater  than 1?

4) If geometric common ratio  is greater than 1 , it means the sum of the series  keep increasing , then we say it is diverges.

5⁴⁰  = 5 * 5³⁹

5 * 5³⁹  -  5³⁹ =

4 * 5³⁹                            which is obviously divisible by '4'         4* x    is divisiable by 4  (x is 5³⁹ in this case)

How many integers between 500 and 1000 contain both the digits 3 and 4?   

Two ways the integer can contain both 3 and 4 ... to wit, it ends with 34 or 43. 

So, in the 500's there are 534 and 543  

similarly there are             634 and 643 

                                         734 and 743  

                                         834 and 843  

                                         934 and 943              total:  10 such integers   

_.

Plugging into the web2.0 calculator:

1+2+3+4+5+6+7+8+9+10+6+14+5+4+9+4+5+5+55+2+2+5+5+5+45+2+5+1000x2 = 2233

11 m/hr * 5280 ft / mi  * 12 in / ft * 1 hr /60 min = 11616 in/min

Circumference circle = 2 pi r  

  13 in * 2 * pi = circumference of wheel per  revolution

11616  in/m  /  26 pi in / rev  =    142.21  rev /min

142.21 rev / min  *   2pi Radians / rev = 893.54  R/min

a = 1/4 (first term)

r = 1/2 (ratio)

n = number of terms

15/32 = 1/4((1/2)^n - 1)/(1/2 - 1)

Try to solve it by yourself from here. :))

15/32 = 1/4((1/2)^n - 1)/(1/2 - 1)

15/8 = ((1/2)^n - 1)/(-1/2)

-15/16 = (1/2)^n - 1

1/16 = (1/2)^n

n = 4

=^._.^=

Sine(x) returns a number from -1 to 1, not an angle. 

=^._.^=

If \(x\) and \(y\) satisfy \((x-3)^2 + (y-7)^2 = 64\),
what is the minimum possible value of \(x^2 + y^2\)?

\(\text{Let the origin $O=(0,0)$ } \\ \text{Let the center of the circle $C=(3,7)$ } \\ \text{Let the radius of the circle $r=\sqrt{64}=8$ } \\ \text{Let $x^2+y^2=r_{\text{min}}^2$ }\)

\(\text{1. Distance between origin and center of the circle:}\\ \begin{array}{|rcll|} \hline \overline{OC} &=& \sqrt{(0 - 3)^2 + (0 - 7)^2} \\ \overline{OC} &=& \sqrt{9+49} \\ \mathbf{\overline{OC}} &=& \mathbf{\sqrt{58}} \\ \hline \end{array} \)

\(\text{2. } x^2+y^2 = r^2_{\text{min}} \\ \begin{array}{|rcll|} \hline r_{\text{min}} &=& r-\overline{OC} \\ r_{\text{min}} &=& 8-\sqrt{58} \\ \hline x^2+y^2 &=& r_{\text{min}}^2 \\ x^2+y^2 &=& \left(8-\sqrt{58}\right)^2 \\ x^2+y^2 &=& 64-16\sqrt{58} + 58 \\ \mathbf{ x^2+y^2 } &=& \mathbf{ 122 -16\sqrt{58} } \\ \text{The minimum value of }~\mathbf{ x^2+y^2 } &=& \mathbf{0.1476303062 } \\ \hline \end{array}\)

AD = 3            [ABC] = 16²/₃           BD = x

(√3x)(3 + x) = 2(16²/₃)

x = 5¹/₃

CD = (2*16²/₃ ) / 8¹/₃ = 4

[ACD] = 6 cm²

[BCD] = [ABC] - [ACD] = 10²/₃ cm²

[ACD] / [BCD] = 9/16 

f(x) = 7x + 5

and g(x) = x + 1

⇒ \(h(x)=f(g(x))\)

              \(=f(x+1)\)

              \(=7x+7+5\)

              \(=7x+12\)

In order to find  \(h^{-1}(x)\) 

Let  \(y=h(x)\)

       \(y=7x+12\)

       \(x= {y-12\over 7}\)

⇒ \(h^{-1}(x) = {x-12 \over 7}\)

   Thus, inverse of h(x) is \({x-12 \over 7}\)

~Amy 

Now in this problem, 

Let the no. of  10 dollar bills be x, 5 dollar bills be y, 2 dollar bills be z and 1 dollar bills be w.

According to question, 

\(10x+5y+2z+w=20\)                    \(...(1)\)

In order to find no. of ways, we've to find no. of solutions eq. (1) has.

∴ By the theory of combinatrics, 

No. of solutions \(=\left( \begin{array}{c} 20+4-1 \\ 4-1 \end{array} \right)\)

                          \(=\left( \begin{array}{c} 23 \\ 3 \end{array} \right)\)

                          \(=1771\)

∴ Eq. (1) has 1771 possible set of solutions.

Thus a $20 bill can be changed in 1771  possible ways. 

~Amy 

According to the question, 

All the integers between 500 & 1000 are 3 digit no.s greater than 500. 

∴ First place may have digits 5,6,7,8 & 9 = 5 possibilities

Second & third place may have digits 3 or 4 = 2 possibilities 

⇒Total no. of possibilities \(=5*2\)

                                         \(=10\)

Thus, total no. of such integers are 10. 

~Amy 

Let's consider this graph 

So as you can see the seismographs are placed at \(A(-2,5), B(2,3)\) and  \(C(-2,-4)\).

From the question, 

Epicentre is located 2 miles from A, 4 miles from B & 7 miles from C

∴ Eliminating all possibilities, epicentre is located at \(E(-2,3)\).

~Amy 

If
\(4x = 3y\),
what is the value of \(\dfrac{2x + y}{3x - y}\)?

\(\begin{array}{|rcll|} \hline 4x &=& 3y \\ \mathbf{ y } &=& \mathbf{ \dfrac{4}{3}*x } \\ \hline \dfrac{2x + y}{3x - y} &=& \dfrac{2x + \dfrac{4}{3}*x}{3x - \dfrac{4}{3}*x} \\\\ &=& \dfrac{ x\left(2+\dfrac{4}{3}\right) }{ x\left( 3 - \dfrac{4}{3}\right) } \\\\ &=& \dfrac{ 2+\dfrac{4}{3} }{ 3 - \dfrac{4}{3} } \\\\ &=& \dfrac{10}{3} \over \dfrac{5}{3} \\\\ &=& \dfrac{10}{3} * \dfrac{3}{5} \\\\ &=& \dfrac{10}{5} \\\\ \mathbf{\dfrac{2x + y}{3x - y}} &=& \mathbf{2} \\ \hline \end{array}\)

Joshua rolls two dice to form a two-digit integer.
If the number on the first die represents the tens digit
and the number on the second die represents the units digit,
what is the probability that the integer formed is divisible by 8?

\(\begin{array}{|c|c|c|c|c|c|c|} \hline & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 11 & 12 & 13 & 14 & 15 & \color{red}16 \\ \hline 2 & 21 & 22 & 23 & \color{red}24 & 25 & 26 \\ \hline 3 & 31 & \color{red}32 & 33 & 34 & 35 & 36 \\ \hline 4 & 41 & 42 & 43 & 44 & 45 & 46 \\ \hline 5 & 51 & 52 & 53 & 54 & 55 & \color{red}56 \\ \hline 6 & 61 & 62 & 63 & \color{red}64 & 65 & 66 \\ \hline \hline \end{array}\)

The probability is \(\dfrac{5}{36}\) or \(13.9~\%\)

nvm. i got it. it was 3: 7, 31, and 41