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\(1 + \frac{3}{6} + \frac{5}{6^2}\)

\(\frac{6}{6}+\frac{3}{6}+\frac{5}{36}\)

\(\frac{9}{6}+\frac{5}{36}\)

\( \frac{54}{36}+\frac{5}{36}\)

\(\mathbf{\frac{59}{36}}\)

Alright, let's note that the sector has an angle of 83 degrees. 

This means the area of the sector is \(83/360\) of the total area. 

We can write the equation 

\(\frac{83}{360}\pi r^2 = 60\pi\\ 83/360r^2=60\\ r^2=60*360/83\\ r^2 = 21600/83\\ r=\sqrt{216000/83} \approx 16.132\)

Thanks! :)

Since all sides of a square are equal, the top side of the square is one. Since the sides of equilateral triangles are also the same, all of the sides on the equilateral triangles are also one. The bases of two of these equilateral triangles almost make up the side length of the large square, but there is a gap. If we continue the horizontal line made up by those two bases, we can see that the shape to the right of the last equilateral triangle is another equilateral triangle, just cut off halfway. This means that the whole length of the side of the square is 2 sides of the equilateral triangle plus one half of the side, which is 1 + 1 + 1/2 = 5/2. To find the area of the large square, we just square the side length, getting (5/2)^2 = 25/4 square units.

If your team  has one win and 11 losses after four weeks....and  then wins one  game and loses two per week after that, you can't ever get to 60% !!!

Since the ratio between succesive terms is the same we  have

[100 +n ] / [ 60 + n]  = [ 140 + n ] / [ 100 + n ]     cross- multiply

[100 + n ] [ 100 + n ]  = [ 60 + n ] [140 + n]

10000 + 200n + n^2  =  8400 + 200n + n^2

10000   = 8400 

No solutions exist

a1 + a3  = 5

a2  + a4 = 6

a2  = a1 + d

a3 = a1 + 2d

a4 = a1 + 3d     where d = the difference  in  successive terms....so....

a1 + (a1 + d)  = 5

(a1 + d) + (a1 + 3d)  = 6         simplify these

2a1 + d = 5         (1)

2a1 + 4d = 6       (2)

Subtract (1) from (2)

3d  = 1

d =1/3

Using (1)

2a1 + (1/3)  =  5

2a1  =  14/3

a1  =  14 / 6  = 7 / 3

Alright. Let's first put y in terms of x. 

We get \(y=25-x\). 

Subbing this value back into the second equation, we get 

\(3x+75\). 

Since x can't be negative, the smallest possible value of x is 0. 

When x is 0, we have \(3(0)+75=75\)

So 75 is our final answer. 

Thanks! :)

You might want to double check the question, because there is no real pair of numbers that holds those properties. There are, however, complex numbers that do:

\(y = 10 - x\)

\(x^3 + (10-x)^3 = 162 +x^2 + (10-x)^2\)

\(x^3 + (10-x)^2(10-x) = 162 + x^2 + (100 - 20x + x^2)\)

\(x^3 + (100-20x+x^2)(10-x) = 2x^2-20x+262\)

\(x^3 + (1000 -200x+10x^2-100x+20x^2-x^3) = 2x^2-20x+262\)

\(30x^2-300x+1000 = 2x^2 - 20x + 262\)

\(15x^2-150x+500=x^2-10x+131\)

\(14x^2-140x+369 = 0\)

\(x^2-10x+\frac{369}{14} = 0\)

\(x^2 - 10x +\frac{350}{14} = -\frac{19}{14}\)

\(x^2 - 10x + 25 = -\frac{19}{14}\)

\((x-5)^2 = -\frac{19}{14}\)

\(x-5 = \sqrt{\frac{19}{14}}i\)

\(x = 5 + \sqrt{\frac{19}{14}}i\)

\(y = 10 - (5 + \sqrt{\frac{19}{14}}i)\)

\(y = 5 - \sqrt{\frac{19}{14}}i\)

Therefore, one ordered pair of complex numbers that works is 5 + sqrt(19/14)i and 5 - sqrt(19/14)i

Please let me know if I had made any errors!

https://web2.0calc.com/questions/algebra_40599

range  = [ -7 , 11 ]

obviously....x cannot  = 0

1 + 3/x   cannot = 0   ...so   1 + 3/x  =  [ x + 3] / x

And    2/ ( [ x + 3] / x)  =  2x / [ x + 3]

So x cannot =  -3

Lastly

1 + [2x] / [ x + 3]   cannot  = 0

[ x + 3 + 2x] /[x + 3]

[ 3x + 3 ] / [x + 3]

So......  1 / [ (3x + 3) / (x + 3) ]

(x + 3 )  / (3x + 3)      

So x cannot = -1/3

\((3-\sqrt{5})(4+\sqrt{5})(1+6\sqrt{5}) \)

First two products =  7 -sqrt 5

So (7 -sqrt 5) (1 + 6sqrt 5)  =

7 -sqrt 5 + 42sqrt 5 - 6*5  =

-23  + 41sqrt 5

a + b =   18

Here's a similar prob

https://web2.0calc.com/questions/truck-problem

If you get stuck, let me  know

\(9^x - 2 \cdot 3^x + 1 \)

(3^x - 1)^2

Take the derivative and  set to 0    {close your eyes if you haven't had Calculus!!   }

2 (3^x -1) ( 3^x * log 3 ) = 0

(3^x -1) * 3^x   =  0        { 3^x  is always +]      [divide this out ]

3^x -1  = 0

3^x  =1

x = 0

The min is

9^0  - 2*3^0 + 1    =     1  - 2(1) + 1   =  0

x^2 + 3y^2   =18

Take the derivative with respect to x and  set to 0

2x = 0

Take the derivative with respect to y and set to 0

6y = 0

This implies that   2x  = 6y

y = (1/3)x

So

x^2  + 3 [(1/3) x]^2  = 18

(4/3]x^2 =18

x^2  = 54/4 

x =  sqrt [54/4]    =  sqrt [27/2]  =  3sqrt (3)  /sqrt (2)  = (3/2)sqrt 6

y = (1/2)sqrt 6

Max is

 x + y =    (3/2)sqrt (6)  + (1/2)sqrt (6)  =    2sqrt 6

Area of  square  (without proof)   =   [ product of leg lengths / sum of leg lengths ] ^2  =

( [ 1 * 3 ] / [ 1 + 3 ] )^2  =  ( 3 / 4)^2  =  9 /16

I can easily prove this if you want me to 

BAC  = 60

So angle BAD  =30    (since AD is a  bisector)

Angle ABC  =45

Therefore

DB / sin BAD  = AD /sinBAD  

DB / sin 30  = 24 / sin 45

DB  / (1/2)  = 24 /sin45

DB = (1/2) * 24 / [sqrt (2) / 2)   = 12sqrt 2

Angle ADB  =  180  - 45 - 30  =  105

So

AB / sin ADB  = AD /sin ABC

AB / sin 105  = 24 / sqrt (2) /2

AB  = 24sin105 /[sqrt (2) /2 ]   = 12 + 12 sqrt (3)

AC / sin ABC  = AB /sin ACB

AC / (sqrt (2)  / 2) = (12 + 12 sqrt (3)) / sin 75

AC = (sqrt (2)  / 2) (12 + 12sqrt (3)) / sin 75   = 24

[ABC ] = (1/2) AB * AC * sin 60

[ABC ] = (1/2) (12 + 12 sqrt (3)) *24 * sqrt (3) /2 =  72 (3 + sqrt 3)  ≈ 340.7

Law of Cosines

BC  =sqrt [BD^2 + DC^2 -2 (BD * dC) cos 60]

BC =sqrt [ 4^2 + 5^2 - 2(4*5) * (1/2) ]  = sqrt 21

Law of Sines

sin BCD / BD = sin BDC / BC

sin BCD =  BD * sin BDC  /BC

xin BCD = 4 sqrt (3)  /2 / sqrt 21

sin BCD  = 2sqrt (3) / sqrt (21)  =  2/sqrt 7

Since BCD and ACE are complementary, then sin BCD = cos ACE =  2/sqrt 7

sin ACE =  sqrt [ 7 - 4] / sqrt 7  = sqrt (3)/sqrt (7)

sin AEC / sin ACE  = AC / AE

sqrt (3) / 2  / [ sqrt (3)/sqrt(7)  = sqrt (7) / 2  = AC /AE

AC = (sqrt (7) / 2) AE

Law of Cosines

AE^2   = AC^2 + CE^2   -2 ( AC * CE) cos ACE

AE^2  = (7/4)AE^2 + 4^2  - 2 ( sqrt (7) / 2 * AE * 4) (sqrt (3)  /sqrt (7)

Let AE = x

x^2  = (7/4)x^2 + 16  - 4sqrt (3) x

(3/4)x^2 - 4sqrt(3)x + 16  =  0

3x^2 - 16sqrt (3)x + 64  = 0 

Solving for x

x = 8 / sqrt 3  =   AE 

This answer happened to be wrong...:(

First, let's calculate how far Will got before Grace even started to row, 

We have \(d=st\), where d is distance, s is speed, and t is time. 

We have \((45 min)(50 m/min) = 2250 m \). Will rowed 2250m before Grace began. 

Now, Will and Grace have to cover \((2800 m) – (2250 m) = 550 m \) together. 

We have 

\( (50)(t) + (30)(t) = 550 \\ 80t = 550 \\ t = 550/80 = 6.875 \)

6.875 minutes is about 6 minutes and 52 seconds. 

The two started at 6:45, so they arrive at about \(6:51:52\) P.M. when they meet. 

Thanks! :)

While it may not seem the most important, let's calculate the semiperimeter. We get \(\frac{3+4+6}{2}=\frac{13}{2}\)

The reaon why we must find the semiperimeter is because we can use Heron's Formula. 

It states that the area is \(\sqrt{s(s-a)(s-b)(s-c)}\) where s is the semiperimeter and a,b, and c are the sidelengths. 

Applying this, we get \(Area = \sqrt{\frac{13}{2}(\frac{13}{2}-3)(\frac{13}{2}-4)(\frac{13}{2}-6)}\)

\(Area = \sqrt{\frac{455}{16}}=\frac{\sqrt{455}}{4}\)

This rounds to about \(5.3327\)

Thanks! :)