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In order to do this, we can set some variables.

Let's set \(x=0.\overline{2123}\)

If x is equal to that, then notice that \(10000 x = 2123.\overline{2123}\)

This is very important to the problem. 

Subtracting x from 10000x, we get

\(\eqalign{10000 X &= &\hfill2123.2123...\cr X &= &\hfill0.2123...\cr \hline 9999X &= &2123\cr}\)

Notice that the repeating decimal cancels out. 

Now, we solve for x. We get

\(X = \frac{2123}{9999}\\X = \frac{193}{909}\)

So our final answer is \(\frac{193}{909}\)

Thanks! :)

First, let's add all three numbers up. We get

\(333+183+401=917\)

Now, we must find factors of 917. 

There's actually only 4 of them, two of them which are 917 and 1. 

The other two are 7 and 131. 

\(7*131=917\)

So our two prime factors are 7 and 131. 

We have \(7+131=138\)

So 138 is our final answer. 

Thanks! :)

First, we put \(0.11012 \) in fraction form not reduced or simplified. 

To do this, we multiply the decimal by \(100000/100000\)

We get \(\frac{0.11012}{1}\times \frac{100000}{100000}= \frac{11012}{100000}\)

Now, we search for common factors. 4 is one. 

\(\frac{11012 \div 4}{100000 \div 4}= \frac{2753}{25000}\)

Unfortuantely, there are no more common factors after this, so we have

\(0.11012 = \frac{2753}{25000}\)

So our final answer is \(0.11012 = \frac{2753}{25000}\)

Thanks! :)

Let's first factor each number to the smallest it can be. We get

\(3 \cdot 41 + 3 \cdot 2^3 \cdot 19+ 3 \cdot 263\)

Notice that every number is a factor of 3. We get

\(3(41+152+263)\)

Adding every number inside the parenthesis, we get

\(3(456)\\ 3^2 \cdot 2^3 \cdot 19 \)

So 3, 2, and 19 are our numbers. 

\(3+2+19=24\)

So 24 is our answer. 

Thanks! :)

We can use variables and equations to solve this problem. 

First, let's let \(x=2.\overline{57}\)

If we have that for the value of x, we have \(100x = 257.\overline{57}\)

This is extremely important for our solution. 

Now, we subtract x from 100x. We have

\(100x-x=257.\overline{57}-2.\overline{57}\)

Notice the repeating decimal cancels out. We get

\(99x=255\\ x=255/99\\ x=\frac{85}{33}\\ x=2\frac{19}{33}\\\)

So our final answer is \(x=2\frac{19}{33}\)

Thanks! :)

We can keep couting up and up until we get prime numbers. 

First, we have \(101\). Notice it's only divisble by 1 and itself, so there's the first prime. 

All the even numbers obviously don't work, so we skip them. 

Next, we have \(103\). It also is only divisble by 1 and itself, so that's the second prime. 

Third, we have 105. It's divisble by 5, so it doesn't work. 

Then, we have \(107\), which is also a prime number. 

\(109\) is also prime, so we count that as our fourth prime number

111 is divisble by 3 and 37, so it doesn't work.

However, \(113\) is prime, so that's the smallest. 

So 113 is our answer. 

Thanks! :)

A prime number is only divisble by 1 and itself. 

First, let's look at 1. 

Even though it is divisible by 1, it only has one true factor, meaning it is not prime. 

Next, we have 2. Two is only divisble by 1 and itself, so it qualifies. 

Following this pattern, \(3=(3*1), 5=(5*1), 7=(7*1)\) are all prime. 

However, \(4 = 2*2, 6=2*3, 8=2*4, 9=3*3, 10=2*5\)are all what we call composite. 

Thus, we have \(2,3,5,7\) as our prime numbers. 

Thanks! :)

A                         B

                           4

                           P

                           1                          

D    2     Q    3    C

PA =  sqrt [ 5^2 + 4^2] = sqrt 41

PQ = sqrt [ 3^2 + 1^2] = sqrt 10

AQ = sqrt [ 5^2 + 2^2]  = sqrt 29

Law of Cosines

PQ^2 = PA^2 + AQ^2 - 2(PA * AQ) cos (PAQ)

10 = 41 + 29  - 2 ( sqrt [41 * 29] ) cos (PAQ)

cos (PAQ)  =  (10 - 41 - 29 ) /  (-2 *sqrt [ 41 * 29 ] )  =  30 / sqrt 1189

sin (PAQ)  =  sqrt [1189  - 30^2 ] / sqrt 1189   = sqrt [289] / sqrt 1189  =  17 / sqrt 1189

Simplify as

x^2 + 12x + y^2 + 2y  =  10        complete the square on x, y

x^2 + 12x + 36 + y^2 + 2y + 1 =  10 + 36 + 1

r^2  =  10 + 36 + 1   = 47

Area  = r^2 * pi =  47 pi ≈ 147.65

What is the greatest prime factor of the sum of the arithmetic sequence 1 + 2 + 3? 

1 + 2 + 3  =  6  

the factors of 6 are 2 x 3  

greatest  =  3  

both 2 and 3 are prime, the larger is 3  

_.

a)  3 (8/10)  + 8 (2/10)  =  24/10 + 16/10  =  40 / 10   =  4 

b) 3(8/11) + 8(3/11)  =  24/11 + 24/11  = 48 / 11  ≈  4.36

c) 3 (8/12) + 8 (4/12)  =   24/12 + 32/12  =  56 / 12 =  14 / 3 ≈ 4.67 

d)  3 [ 8 / (10 + n) ]  +  8  [ (2 + n) / (10 + n)  ]  = 6

[ 24 + 16 + 8n ] / [10 + n ]  =  6

40 + 8n  = 6 [ 10 + n]

40 + 8n  = 60 + 6n

8n - 6n =  60 - 40

2n = 20

n = 10

e)   3 [ 8 / (10 + n) ]  +  8  [ (2 + n) / (10 + n)  ]   ≥ 7

40 + 8n   ≥  7 [ 10 + n ]

40 + 8n  ≥ 70 + 7n

8n - 7n  ≥ 70 - 40

n  ≥ 30

x^2 -10x + y^2 + 10y  =  -c     complete the square on x, y

x^2 -10x + 25 + y^2 + 10y + 25  = -c + 25 + 25

-c + 25 + 25   =  1

-c + 50  =1

-c = -49

c = 49

We have 2 Americans and we can choose one of them to receive a medal. This can be done in 2 ways.

Choosing a Canadian to receive a medal:
We have 5 Canadians, and we can choose one of them to receive a medal. This can be done in 5 ways.

Choosing a third skater to receive a medal:
After selecting an American and a Canadian, we are left with 17 skaters (19 - 1 American - 1 Canadian). This can be done in 17 ways.

Thus, we have:2 (American's choice).×5(Canadian's choice)×17(third skater's choice)=170 ways
 

Let x be the smallest angle

Let  x + 2d   = the largest

(x + 2d)  - x  = 56

2d = 56

d =28

So

x + (x + d)  + (x + 2d)   =180

3x + 3d   =180

x + d  = 60

x + 28 = 60

x =60 - 28 =  32  = the smallest angle

We will have 6 congruent triangles with sides of 3 and an included angle of 60°

Area  = 6 (1/2) (3^2) * sin 60  =

6 (1/2) (3^2) * sqrt (3)  / 2

27 * sqrt (3) / 2  =

13.5 *sqrt (3)  ≈ 23.4

Draw DE , CF  perpendicular to  AB

Let AE  = x

Let BF = 96 - (44 + x)  = 52 - x

Angle DAB = 70

tan DAB  = h /AE

tan(70)  = h/ x

h = x tan 70

tan (55)  = h / (52 - x)

h = (52 -x)tan (55)

So

x tan 70  = (52 -x) tan 55

Solving  for x   ≈ 17.785

h = x tan 70  =  17.785 tan 70

Area  =  (1/2)  17.785 * tan 70 *( 96 + 44) ≈ 3420 .5

What is DE ?

\(\overline{BC}^2=\overline{AB}^2+\overline{AC}^2-2\cdot \overline{AB}\cdot \overline{AC}\cdot cos\ \alpha\\ \alpha =acos\ ( \dfrac{\overline{AB}^2+\overline{AC}^2-\overline{BC}^2}{2\cdot \overline{AB}\cdot \overline{AC}})\\ \alpha =acos\ (\dfrac{13^2+19^2-3^2}{2\cdot 13\cdot 19})=acos\ (1.0546\ ...)\\ \alpha =\ unreal\)

The segments 3, 13 and 19 do not form a closed triangle.

 !

If a polygon has 3 sides, then it has to be a triangle. 

We can solve this problem by setting some variables. Let's set the smallest angle degrees as n. 

Now, since the angles from an arithmetic sequence, each angle is increasing an equal amount. 

This means if the first and last angle are 56 degrees apart, then the middle angle is \(56/2=23\) degrees bigger than n. 

Thus, we can write the equation \(n+(n+23)+(n+56)=180\)

Solving this equation, we get

\(3n+79=180\\ 3n=101\\ n=101/3\)

So our final answer is 101/3

This is not a whole number, but it does work. 

Thanks! :)

First, we need to combine all like terms to see what we area dealing with. We have

\(x^{2}+18x+3\). 

Now, completing the square for x by adding and subtracting 81, we get 

\((x+9)^2-78\)

Thus, we can easily fill in the blanks now. 

The first blank is 1

The second blank is 9

The third blank is -78. 

Thanks! :)

We can set up an equation to help us solve this problem. Let's define a variable. 

Let n be the smallest  number. We have \(n + n+1 + n+2 ....... n+6 = 7n + 21\)

Now, this sequence adds up to 28/5 times the largest number, n + 6, so we can write the equation

\(7n+21=28/5(n+6)\)

From here, we just solve for n. We have

\(7n+21=\frac{28}{5}n + 168/5\\ 7/5n=63/5\)

Now, multiplying both sides by the reciprocol of 7/5, we get

\(n=\frac{63}{5} \cdot \frac{5}{7}=9\)

So 9 is our final answer!

Thanks! :)

First, we have to find the values of PA, PQ, and AQ.

Using the pythagorean theorem for all 3 of them, we get

 \(AP = \sqrt{4^2+5^2}=\sqrt{41}\)

\(PQ = \sqrt{1^2+3^2}=\sqrt{10}\)

\(AQ=\sqrt{1^2+5^2} = \sqrt{26}\)

Now, we apply the Law of Cosines. According to the law, 

\(PQ^2 = AQ^2 + PA^2 - 2(AQ * PA) * cos (PAQ)\)

Plugging in all the values of AP, PQ, and AQ we figured out, we get

\(10=26+41-2(\sqrt{26}*\sqrt{41})*cos(PAQ)\\ 10=67-2\sqrt{1066}*cos(PAQ)\\ cos(PAQ) = \frac{-57}{-2\sqrt{1066}}\)

Now, sin(PAQ) is just 1 - cos(PAQ), so we get

\(sin(PAQ)=1- \frac{-57}{-2\sqrt{1066}}\\ sin(PAQ)=1+ \frac{57}{-2\sqrt{1066}}\\ sin(PAQ) = \frac{-2\sqrt{1066}+57}{-2\sqrt{1066}}\)

This is approxaimtely \(0.12709645414\)

So our answer is \(\frac{-2\sqrt{1066}+57}{-2\sqrt{1066}}\)

Thanks! :)

Let's first convert both numbers into base 10 before adding them together. We have

\((A710)_{16} = (10 × 16^3) + (7 × 16^2) + (1 × 16^1) + (0 × 16^0) = (42768)_{10}\)

\((51)_{16} = (5 × 16^1) + (1 × 16^0) = (81)_{10}\)

Now we have to add up 81 and 42768. We get \(42768+81=42849\)

Now we want to convert 42849 into base 16. 

We go in order, so we have

\(42849/16 = 2678 R1\)

\(2678/16 =167 R6\)

\(167/16 =10R 7 \)

\(10/16 =0R 10 \)

So our final answer is \(A761\)

Thanks! :)

First, let's set a variable. Let's set x as angle PQR. 

From the problem, we can write that 

\(cos(x) = \frac{8^2 + 11² - 5²}{2*8*11}=10/11\)

Now, we can sue the law of cosines to write

\( (PM)² = 5.5² + 8² - 2*5.5*8 * cos(x)\\ \)

Now, we can find PM. 

\(PM^2 = 10/11*6.25\\ PM \approx 2.38365647311\)

I'm not exactly sure if this is correct. I may have made a mistake. 

Thanks! :)