All Answers 356332 Answers

https://web2.0calc.com/questions/counting-problem_70639

We can easily set up an equation to do this problem. 

We can do

\(\Delta EFG \cdot \frac{2h}{3}=18\\ \Delta EFG =\frac{3\cdot 18}{2h}\\ V_{EFGH}=\Delta EFG\cdot \frac{1}{3}h\\ [EFGH]=\frac{3\cdot 18}{2h}\cdot \frac{1}{3}h\\ [EFGH]=9\)

So 9 is our answer!

Thanks! :)

Let's first notice something really important. 

The list \(P_1 P_2 + P_2 P_3 + P_3 P_4 + \dots + P_9 P_{10} + P_{10} P_1\) are just the sides of a regular decagon!

The sidelength of one side the decagon can be represented as \(\frac{\text{radius}}{2 ( -1 + \sqrt{5})}\)

Plugging in all the values we know, we have

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

So about 6.18 is our answer!

Thanks! :)

Let's first move all terms to one side of the equation and combine all like terms.

We get

\(x^2 - 21x + y^2 + 13y = 25\)

Now, let's complete the square for x and y on the left side the equation. 

\(x^2 - 21x + 441/4 + y^2 + 13y + 169/4 = 25 + 441/4 + 169/4\)

\((x - 21/2)^2 + (y + 13/2)^2 = 355 / 2\)

This is the equation for a circle. 

According to the circle rules, this circle has a center at \((21/2 , -13/2)\)and radius \(\sqrt{355 / 2}\)

The largest possible value is basically the radius added onto the x coordinate of the center. 

We have \(x = (21/2) + \sqrt{355 / 2}\)

Thanks! :)

Let's note something important really quickly. 

If we have \(XWZ = 60°\), then we must have \( WXY = 180° - 60° = 120°\) since XY and WZ are parralels. 

\(\angle ZXW\) has to be less than 120 degrees, but in the problem, it clearly states it is 120 degrees. 

This is an invalid statement, so therefore, this problem is not solvable. 

Thanks! :)

We want to put \(\log_{8a} b\) in terms of a and b. 

It's actually not that hard. 

We just have

\(\frac{ log(b)}{ log (8a)} = \frac{log( b)}{ log (8) + log (a) }\)

That's the best I could do.

Thanks! :)

It's actually not hard to just list all the numbers out. 

We have

\(\frac{1}{3}, \frac{2}{3}, \frac{4}{3},\frac{5}{3}, \frac{7}{3}, \frac{8}{3}, \frac{10}{3}, \frac{11}{3}, \frac{13}{3}, \frac{14}{3}\)

The ones we excluded from this list, like 3/3 and 6/3 are not in simplest form, therefore eliminated. 

Now, adding all these values up, we get \(75/3=25\)

That means 25 is our answer!

Thanks! :)

First, let's calculate the ones inside the parenthesis. 

We have \(16*5*\frac{1}{10} = 8\)

Now, we just have to find \(\log_{1/2} (8)\)

Using logs and bases, we have \(\log_{1/2} (8) = -3\)

So -3 is our answer!

Thanks! :)

There's actually only one triangle that satisfies all the rules stated.

The only one that works is 

2-2-1

Any other triangle would violate the triangle inequality theorems, therefore being invalid. 

So 1 is our answer!

Thanks! :)

First, let's note that PXQ and RXS are similar triangles. 

The scale factor of the two is \( \sqrt{8/4} = \sqrt{2}\)

This means that the height of RXS is \( \frac{h\sqrt{2}}{ 1 + \sqrt{2}}\) where h is the height of the trapezoid. 

We have \(\frac{ h}{1 + \sqrt{2} }\) for the height of PXQ

The base of PXQ is \(\text{base of RXS}/\sqrt{2}\)

Now, using this information from PXQ, we can write

\(\text{Area} = \frac{(1/2) (\text{base of RXS})}{\sqrt{2} * h ( 1 + \sqrt{2})} \\ 4 = (1/2) (\text{base RXS}) * h / ( 2 +\sqrt{ 2}) \\ 8(2 + \sqrt{2}) / \text{base RXS} = h\)

We know the area of the trapezoid is \((1/2) h (\text{sum of bases})\)

We have 

\( 4 [ 2 +\sqrt {2}] [ 2 + \sqrt {2} ] / / 2 \\ 2 [ 2 + \sqrt {2}] ^2 \\ 2 [ 4 + 4\sqrt {2}+ 2] = \\ 2 [ 6 + 4\sqrt {2}] = \\ 12 + 8\sqrt {2} \)

Thanks! :)

xy = 2/5

y = (2/5)x^(-1)

6x +  3  / (5 * (2/5) x^(-1))  =  

6x +  (3/2)x  

(15/2) x

There is no minimum......this is a line through the origin with a slope of  15 / 2

Construct a circle at B  with radius BC  that intercepts the semi-circle at E

Because BC = BE......then BE will be tangent to the semi-circle

Carry BE through to A ....let AD = x

OE = OC

BC = BE

BO = BO

Then triangles EOB and COB are congruent right triangles

tan angle OBC  = 1/sqrt 3  =  tan angle OBE 

tan ( OBC + OBE)  =  tan (ABC)....so.....

[ 2 * 1/sqrt 3 ] /  [1 - (1/sqrt 3)^2]  =   (2 + x) / sqrt 3

2/sqrt 3 * sqrt 3  =  ( 1 -1/3) (2 + x)

2  =  (2/3) (x + 2)

3 = x + 2

x =1

So

AB  = sqrt [ AC^2 + BC^2 ] = sqrt [ (2 + x)^2  + (sqrt3)^2 ]  = sqrt [ 3^2 + 3 ]  = sqrt 12  =  2sqrt 3

   PQ = 36, PR = 22, QR = 26 .....   This is all we need to know   

→  semi-perimeter  =  [ 36 + 22 + 26 ] / 2  = 42....using Heron's Formula

[ PQR]  =   sqrt [  42 *  6 * 16 * 20 ]     = 48 sqrt 35  ≈  284   

Take the derivative  with respect to x and  set it = 0

2x -16  = 0

x = 8

Take the derivative with respect to y and  set to 0

2y + 16  = 0

y = -8

The minimum is

(8)^2 + (-8)^2 - 4(8) + 2(-8) - 12(8) + 14(-8) + 26 =   -102

The slope of a perpendicular line  to y  = 3x + 1 through (7,1)  =  -1/3

The equation of this  line is   y = (-1/3)(x - 7) + 1  = (-1/3)x + 7/3 + 1   =  (-1/3)x + 10/3

To find the intersection point of these two lines

3x + 1 =   (-1/3)x + 10/3

3x + (1/3)x  = 10/3 -1

10/3 x = 7/3

10 x  = 7

x = 7/10 = .7

y = 3(7/10) + 1  =  31/10 = 3.1

The reflected point is   ( .7 - (7 - .7) , 3.1 - (1 - 3.1) )  = ( 1.4 (-5.6 , 5.2)

2. Lateral surface area  =   base perimeter * heighr  =

( 8 + 10 + 12 + 14 + 16) * (12)  =

60 * 12  =

720 mm^2

W      y    A       2        X

                                  x                            

                                  C

                                  x

Z      B        4             Y

angle ACX  + angle BCY =  90

Let

sin ACX = 2 / sqrt [2^2 + x^2]

cos ACX = x/ sqrt ;2^2 + x^2]

sin BCY = 4 /sqrt [4^2 + x^2]

cos BCY = x /sqrt [ 4^2 + x^2]

sin ( ACX + BCY) = sin 90

sin ACX cos BCY  + cos ACX sin BCY

2 / sqrt [ 2^2 + x^2[ * x / sqrt [ 4^2 + x^2]  +  x/ /sqrt [ 2^2 + x^2 ] *  4 /sqrt [ 4^2 + x^2]   =1

2x + 4x    =   sqrt [ 4 + x^2] sqrt [ 16 + x^2]           

6x  =  sqrt [ (4 + x^2) (16 + x^2)]                 square both sides

36x^2  = (4 + x^2) (16 + x^2)

36x^2 = x^4 + 20x^2 + 64

x^4 - 16x^2 + 64  = 0

(x^2 - 8)^2  = 0

x^2 - 8   = 0

x = sqrt (8)

2x = 2sqrt 8

Draw yB  perpendicular to ZY

yB  = 2sqrt 8

yA  = BY - AX  = 4 - 2  = 2

By the Pythagorean Theorem

yA^2  + yB^2  = AB^2

2^2  + (2sqrt 8)^2  = AB^2

4 + 32 = AB^2

36 = AB^2

AB =  sqrt 36  =  6

W                            X

Z      A        B          Y

         4             2

Triangle  ZWB  is a 30-60-90  right triangle

Angle ZWB  = 60

Angle WBZ = 30

So WZ = BZ/sqrt 3  = 4 /sqrt 3

Area of WXYZ =  ZY * WZ  =  6 * 4/sqrt 3  =  24/sqrt 3 =  8 sqrt 3

Let's set the center of the circle as O. 

Draw a line from O perpendicular to the chord. Let this line be OM. 

\(OA = 15 \\ AM = AB / 2 = 4/2 = 2\)

Triangle AOM is right, with angle AMO being 90 degrees.

By using the pythagorean theorem, we have

\(OM = \sqrt{ OA^2 -AM^2} = \sqrt{ 15^2 - 2^2 } = \sqrt{221}\)

So sqrt221 is our answer. 

Thanks! :)

inradius = area  / semi-perimeter

Let AB = BC  = x

semi-perimeter  = [ 20 + 2x ] / 2 =  10 + x

area =  sqrt [ (10 + x) (10 + x -20) (10 +x -x)^2]  =  sqrt [ (10+x) (x -10) (100) ]

5 =   sqrt [ (10+x) (x -10) (100) ] / [ 10 +x]

5 [10 + x] = sqrt [ 100 ( x^2 - 100) ] 

5 [ 10 + x] = 10 sqrt [ (x^2 - 100) ]

[10 + x] = 2 sqrt [x^2 -100]        square both sides

x^2 + 20x + 100  =  4x^2 - 400

3x^2 - 20x - 500 =  0

x = 50/3

Perimeter  =  2 (50/3) + 20  =  160 / 3

Here's a similar problem :  https://web2.0calc.com/questions/help-pls_105

Well, let's first note something. 

We have \(\frac{x}{x} = 1\) as long as x is not 0. 

We have \(\frac{y}{y} = 1\) as long as y is not 0. 

We have \( \frac{z}{z} =1\) as long as z is not 0. 

We finally have \(\frac{xyz}{xyz} = 1\) as long as x,y,z are not 0. 

So, we have \(1+1+1+1 = 4\) as long as \(x,y,z \neq 0\)

So there is only one possible value. 

Thanks! :)

The resuting  solid will equal the difference  in volumes of the cones formed by the rotation of ADC - the rotation of  ADB   =

pi/3 * AD^2  *( DC - DB)  = 

pi/3  * 7.94^2 * ( 9 - 1)  ≈  528.15