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https://web2.0calc.com/questions/find-the-largest-x-value-at-which-the-graphs-of-f

Help Cphill

Let  Izzy's age  =   A

So...Ben's  age  =  4A

And also   ...   A + 6  = 4A

So  

A + 6   = 4A     sbtract A  from both sides

6 =  3A          divide both sides by  3

2  = A

[BTW.....Ben's age  =  8  ]

Thank You soooooooooo Much Geno3141

1/x  >  x     

Note  that there  are  no positive integers, x, that make this true

To  convince yourself.....let   x  =1

Then

1/1  < 1

1 < 1     isn't true

And  if  x  is a positive integer   ≥ 2....the left  side will  be  a fraction beween 0 and 1 , but the right side will always  be greater than this

Quick!

Let rhombus side be       c = 4

1/2 of short diagonal       a = 1.45

1/2 of long  diagonal        b = ?

We have the right triangle  

Larger angle  β = ?                    cos(β/2) = a / c =68.74619°

                                                              β = 137.4923816°     

IT IS TOTALLY______

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im really happy im understanding these! thank you so much your explaination !

New  median   =

[ 29*75  +  23 + 26 + 26 + 30 + 33 + 40 ]   /   [  75  + 6   ] ≈  29.05 

5*6*5*4*4*3*2*1=5*6!*16=57600

We  can use  the Law of Cosines to  find the measure of the smaller angle...we have that

2.9^2  = 4^2 + 4^2  - 2 (4 * 4)  cos  (smaller angle in the rhombus)

Simplify

[ 2.9^2  -4^2 - 4^2 ]  /  [-2 (4 * 4)]  = cos (smaller angle in the rhombus)

Using the arccos  we  have  that

arccos ( [ 2.9^2  -4^2 - 4^2 ]  /  [-2 (4 * 4)] )  =  smaller angle in the rhombus ≈  42.51°

The larger angle in the rhombus  will be supplemental to this = 137.49°  = 137.5°

LOOKS LIKE YER ON A ROLL  !!!

THanks

Area of trapezoid is height x  average of two bases 

    38  *  (24+x)/2 = 581.4     Solve for 'x'   

Pay close attention, here

The triangle  on the left  is a 30-60-90   right triangle

The  sde opposite the 30°  angle  =   1

The side opposite the right angle [ the dashed line]  =   twice this  =  2 

And  the   triangle on the  right is also a 30-60-90  right triangle

The  angle  at the top right  vertex is the 30°  angle  in this triangle

And  the side opposite this = the dashed line  =  2

And  "x"  is opposite side  to the  right angle in this triangle so it's  twice as long as the dashed line =   4

awesome! thank you

THANKS AGAIN

Diagonal   of a square =  side length * √2

No prob......!!!!

THANK U SO MUCH

Triangle ABC is isoceles  with AB  = AC

So   y   =     [  180  - 24  ]   / 2   =    78°

thank u so much

Connect   WY

Then triangle  WXY  is an isosceles  right triangle with leg WX  = leg XY

Using the  Pythagorean Theorem, we  can find  WY  as

√[  WX^2 + XY^2  ]   = √ [4^2 + 4^2 ] =  √32  =  4√2

And since  WX  = XY  then  angle WYX  =45°

So in triangle WYZ, angle  WYZ  = 135  -45  =  90°

So.....triangle  WYZ is also right WY and YZ  being legs and WZ the hypotenuse

So

√ [ WZ^2  - WY^2 ]  =  YZ

√[9^2 -32]  =  √[81 - 32]  = √49  = 7  =  YZ  =  "a"

line 1   re-arranged    y = -3/4 x -14/4       slope, m,  is   - 3/4     perpindicular slope = - 1/m  = 4/3

line 2     y = 4/3 x +b    put in point given

             7 = 4/3 (5) + b     resulting in b = 1/3

y = 4/3 x  + 1/3           m = 4/3    b = 1/3     you can add them....

Draw  perpendicular AB  from  the  top left vertex  to the base

Let  C   be the  bottom left  vertex of the trapezoid

Then  triangle  ABC  is a right triangle  with  AC  the hypotenuse and  BC  a leg

Note  that  BC   =  8  - 5    =  3

So....the  missing side  [ the height of the trapezoid ]  is

√[ AC^2  - BC^2]   = √[7^2  - 3^2] =  √[49 - 9 ] =  √40  =   2√10