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11₂  =  3

10101₂ = 21

11₂ = 3

1100101₂  = 101

1001₂ = 9

1001101₂  = 77 

3 + 21 + 3 + 101 + 9 + 77  = 214

214 / 2  =  107 R 0

107 / 2  = 53 R 1

53  / 2 = 26 R 1

26 / 2 = 13 R 0

13 / 2  = 6 R 1

6  / 2  = 3 R 0

3 / 2 = 1  R 1

1 / 2   = 0 R 1

11010110₂

https://web2.0calc.com/questions/help-meee_22

https://web2.0calc.com/questions/sum_93

We can use a clever trick to solve this problem. 

We have two equations. Subtracting the first equation from the second equation, we get

\((a_2 - a_1) + (a_4 - a_3) + (a_6 - a_5) + (a_8 - a_7) + (a_{10} - a_9) + (a_{12} - a_{11}) = 0\)

The reason why this is important is because in every parenthesis, it forms the common difference d. 

Thus, we have 

\(6d=0\\ d=0\)

This means every number in the sequence is the same. 

Since the entire sequence is equal to 0, we have \(a_1 = 0\)

So our answer is 0. 

Thanks! :)

Let's set variables in order to solve this problems.

Let's let c be the constant added to each number. 

We get \(60+c, 120+c, 160+c\)

Since all the terms form an geometric series, we can write the formula

\(\frac{60+c}{120+c} = \frac{120+c}{160+c}\)

Now, we simplfy solve for c. 

\((c+160)(60+c)=(c+120)(120+c)\)

Distributing everything, combining all like terms, and moving everything to one side, we get

\(-20c-4800=0\)

\(c=-240\)

We plug this back in, and we get the series \(-180, -120, -80\)

This has a common ratio of \(180/120=3/2\)

So 3/2 is our answer, 

Thanks! :)

First, let's make some observations. 

Note that all points that are 5 units away from the point (2, 7) froms a circle with radius 5 and center (2, 7). 

The equation for that circle would be \((x-2)^2+(y-7)^2=25\)

Combining this with the second equation, we get a system. 

\((x-2)^2+(y-7)^2=25\\ y=5x-28\)

We already have a y value in terms of x, so we plug that into the first equation. 

We get

\((x-2)^2+(5x-35)^2=25\\ x^2-4x+4+25x^2-350x+1225=25\\ 13x^{2}-177x+602=0\)

Now, we can solve for x. Using the quadratic equation, we get

\(x=\frac{-{(-177})\pm \sqrt{{(-177)}^{2}-4\cdot {13}\cdot{602}}}{2\cdot {13}}\)

\(x=\frac{177\pm 5}{26}\)

\(x=7\\ x=\frac{86}{13}\)

Plugging these values for y, we get 

\(\begin{pmatrix}x=7,\:&y=7\\ x=\frac{86}{13},\:&y=\frac{66}{13}\end{pmatrix}\)

So our final answer is (7, 7) and (86/13, 66/13)

Thanks! :)

https://web2.0calc.com/questions/circles_147

A =10     E  =  14

B  =11    F   = 15

10 * 16^3  + 7 * 16^2  + 1 * 16 + 0   = 42768

5 * 16 + 1  =  81

11* 16^3  + 8 * 16^2  + 10 * 16 + 14  = 47278

42768 + 81 + 47278 = 90127

90127 / 16  = 5632 R 15

5632 / 16  = 352 R 0

352  /16  = 22 R 0

22/16  = 1 R 6

1/16  = 0 R 1

1600F₁₆

Let the first term  = 2

Let the last term =  2 + 5(n-1)  =  5n - 3

Sum  =  [ 2 + 5n -3 ] [n/2]  > 100

[ 5n - 1 ] [ n ] > 200

5n^2 - n > 200

5n^2  - n - 200 > 0

Set this = 0

5n^2 - n -200 =  0

n ≈ 6.4253

n = ceiling [ 6.4253 ]  =  7

The 7th term produces the scream  =  5(7) - 3  =  32

2,  7,   12,  17,   22 ,   27 ,  32

Sum  9    21   38    60     87     119

2(3)^2 + 2(3) + 1 =  25

1(3)^2  + 2(3) + 2 =  17

2(3)^2 + 0 (3) + 2  = 20

1(3)^2 + 1(3) + 0  = 12

25 + 17 + 20 + 12 =   74

74 / 3  =  24 R 2

24 / 3  = 8 R 0

8 /3  =  2 R 2

2/3  = 0 R 2

2202₃

3(8)^2 + 5(8) + 7   = 239

2(8)^2  + 2(8) + 3 = 147

1(8) + 1  =  9

239 * 147  * 9   =  316197

316197 / 8  = 39524 R 5

39524 / 8  =  4940 R 4

4940 / 8  = 617 R 4

617 / 8  = 77 R 1

77/8  = 9 R 5

9 /8  =1 R1

1/8  = 0 R 1

1151445₈

90₁₁  =  9 (11) + 0 = 99

So

B  = 11

13

14  could be arranged as 7 x 2

15 could be arranged as 5 x 3

13 can only be arranged as 13 x 1

a3  = a1 + 2d

a2 = a1 + d

a4 = a1 + 3d

a6 = a1 + 5d

So we have these equations

a1 + a1 + 2d = -2

a1 +d + a1 + 3d + a1 + 5d = 1     simplify

2a1 + d = -2

3a1 + 9d = 1

Multiply the first equation by -9  and add to the  second

-15a1 = 19

a1  =  -19 / 15

This is impossible

Leg AC  would be >  6  while  AB (the hypotenuse)  would only  = 1

x^2 + 5y^2  = 30

Take the derivative with respect to x and set to 0

2x = 0

Take the derivative with respect ot y and  set to 0

10y = 0

This implies that

2x =10y

x = 5y

So

(5y)^2 + 5y^2   =30

30y^2  = 30

y = 1

x = 5y  =  5

Mav  x + y   =  6

In order to solve this problem, we have to note something very important!

\(P_1 P_2 P_3 \dotsb P_{10}\) forms a regular decagon!

This means that \(P_1 P_2 + P_2 P_3 + P_3 P_4 + \dots + P_9 P_{10} + P_{10} P_1\) is just the perimeter of that regular decagon

In order to find one side length of the decagon, we can use the equation 

\(\frac{\text{radius}}{2( -1 + \sqrt {5})}\)

Using information from the problem, we can easily find the perimeter.

 \(10 (1/2) ( -1 + \sqrt{ 5}) = 5 ( -1 + \sqrt{ 5}) ≈ 6.18\)

So 6.18 is about our answer!

Thanks! :)

Let's set some variables and make some observations. 

Let's let \(d = a_2 -a_1\)

Now, let's note something important. Note that

\(S_5 = 5a_1 + (d + 2d + 3d + 4d) = 5a_1 + 10d\)

\(S_{10} = 10a_1 + (d + 2d + \cdots + 9d) = 10a_1 + 45d\)

\(S_{15} = 15a_1 + (d + 2d + \cdots + 14d) = 15a_1 + 105d\)

These observations are really important for us to solve the problem. 

Let's take two of these cases. We can write a system of equations, and we get

\(\begin{cases} 5a_1 + 10d = \dfrac15\\ 10a_1 + 45d = \dfrac1{10} \end{cases}\)

Now, we solve for a1 and d. 

Solving this system of equations, we get

\(d = -\dfrac3{250}\\ a_1 = \dfrac8{125}\)

Plugging this into S15, we get

\(S_{15} = 15a_1 + 105d = -\dfrac3{10}\)

So -3/10 is our final answer. 

Thanks! :)

Using the information from the problem, we can write an arithmetic sequence. 

For that sequence, let's let n be the number number in the sequence. We can write the expression

\(2+5(n-1)\)

We want this number to be greater or equal to 100, so we can write the inequality

\(2+5(n-1) \geq 100\)

Now, we solve for n, 

\(2+5n-5 \geq 100\\ -3+5n \geq 100\\ 5n \geq 103\\ n \geq 20.6\)

Since n has to be an integer, the smallest n can be is 21. 

This means that n is 21. Plugging this back in, we get

\(2+5(21-1)\\ 2+5(20)\\ 2+100\\ 102\)

So our final answer is 102 

Thanks! :)

xy – (x + y) = 39    

when x and y are relatively prime and less than 2  

I'm going out on a limb and saying there aren't any.  

If we hypothesize that the problem meant less  

than 20, and the 2 is a typo, then (11,5) works. 

If we can ignore the non-prime and the limit of 2,    

then two more pairs that work are (6,9) and (21,3).  

They were pretty easy to find by graphing it on Desmos.   

_.

First, let's add up all 3 of the numbers we have. We get

\(5005 + 6192 + 11823=23020\)

Now, we want to find the prime factors of 23020. We know that 23020 is divisble by 10, so 2 and 5 are factors. 

\(23020/10=2302\)

2302 is divisble by 2, so we have \(2302/2=1151\)

1151 is actually prime, so we can stop here. 

This means that 23020 has 3 prime factors. \(1151, 2, 5\)

Adding all of these on, we get

\(1151+2+5=1158\)

So 1158 is our answer. 

Thanks! :)

Let's first add up all 3 of the numbers. We get

\(781 + 691 + 3858=5330\)

Now, we have to find the prime factors of 5330. Notice that 5330 is divisble by 10. This means that 2 and 5 are factors. 

\(5330/10=533\)

533 has 2 factors other than 1 and itself. The two are \(13, 41\)

This means that 5330 has 4 prime factors.

\(2, 5, 13, 41\)

Adding these up, we get

\(2+5+13+41=61\)

So 61 is our answer. 

Thanks! :)

First, let's add all 3 of these numbers up. We get

\(728+118+2024=2870\)

Now, we want to find the prime factors of 2870. 

2870 is divisble by 10, so we have 5 and 2. 

\(2870/10 = 287\) 287 only has 2 factors other than 1 and itself, which is \(7, 41\)

This means that we have 4 prime factors. \(5, 2, 7, 41\)

Adding these up, we get

\(5+2+7+41=65\)

So 65 is our answer. 

Thanks! :)

First, let's add up all 3 of the numbers. 

We get \(248 + 113 + 2870 = 3231\)

Now, we have to factor 3231. Notice that \(3+2+3+1=9\), so 3231 is divisble by 3. 

We get \(3231/3=1077 \). Notice that \(1+7+7=15\), so 1077 is also divisble by 3. 

\(1077/3 = 359\)

359 is prime. 

So now, we have 2 factors, 3 and 359. 

\(3+359=362\)

So our final answer is 362. 

Thanks! :)