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Take 1/2 of 42  = 21

Square it   = 441

k =  441   -16   =   425

Proof

y^2 +  42y  + 16 + 425   = 

y^2  + 42y +  441    =

(y + 21)^2 

Good job, Spring  !!!!

Interior angle of  tegular  hexagon =   (6-2)(180)/ 6  =  (2/3) 180 =  120  = BAD

Interior aangle of regular pentagon  =  (5-2) (180)/5 =  108  = CAD

BAC  =  BAD - CAD  =  120  -  108   =   12°

Corrected  !!!

1. True: 210° = 7π/6

2. False: 120° = 2π/3 (not 150°)

3. True: 270° = 3π/2

Hope this helps! :)

sin A  =   BD/BA   =1/2

BD / BA   =1/2

BD  =(1/2) BA

BD  = (1/2)24  =   12

sin C  =  BD / BC  = 1/4

BD/ BC   =1/4

BC /BD  =  4

BC = 12 (4)   =  48

4π/3 = 240°

5π/4 = 225°

2π = 360°

Hope this helps! :)

B.) The sine of an angle is defined as the length of the side opposite over the length of the hypotenuse. (This is the correct answer).

An easy way of remembering the trigonometric ratios is by memorizing "sohcahtoa."

Sine = opposite/hypnotenuse (soh)

Cosine = adjacent/hypotenuse (cah)

Tangent = opposite/adjacent (toa)

Hope this helps! :)

(x + 7) ( x - 1)   = 0

x^2   + 7x  - 1x     -  7   =   0

x^2 + 6x  - 7     =   0

The interior angles sum to 540

So.....  just solve this for  x   and it will  be your  answer

90 * (2)  + 3x   =  540

Since   l2 is paralllel  l1    and m2 is perpendicular to l2 , then  m2  is also perpendicular to m1

Then the  angle  lying  below  1   =  90 - 50  =  40

And this  equals  the angle below angle 2

But this angle  and angle 2  are supplementary

So

40  +   angle  2   = 180

angle 2   =  180   - 40   =    140°

Note  that   if we  connect BD  we have square  ABDE  with a side  = 4

So...its area =  side^2 =  4^2  = 16

y coordinate of  midpoint of BD =    4 

Then  the area of  triangle BDC =   50  - 16  =  34

So....the  area of this triangle = (1/2)  base  * height =  

(1/2) (BD) (height)   =   

(1/2)  (4) height  =   34       

2 * height =  34

height =  34/2    = 17

So.... C =   (2, y midpoint of BC + 17 )  =    (2, 4 + 17)   =  (2 , 21)

In  a 30 -60 -90   right triangle, the  hypotenuse =  2/√3   times as long as the side  oppositie the 60° angle

So      ....        9 *2/√3 =    18/√3   =   18√3/3    =   6√3

4x + 7y +  c   =   0        Re-write  as

y =  -(4/7)x  +  c / 7

y intercept....let  x  =   0

y =  c/7

x int  ....let  y   = 0

0 = (-4/7)x + c/7

(4/7) x  =  c/7

4x  = c

x = c/4

So

c/4  + c/7    = 24

(11/28) c   =24

c =  (28/11) (24)

c = 672/11

Easy  way to  do this

68  +  38/60  +  11/3600   = 

68.636°

-11pi/12  =  2pi - 11pi/12  =  [ 24pi - 11pi ]  / 12  =   13pi/12  =  pi + pi/12  =     pi/12 

-23pi/6  =  -3pi  -  5pi/6  =   pi  - 5pi/6  = 180 - 150  = 30  =   pi/6

7pi/6 =  pi + pi/6  =   pi/6

7pi/4  =   pi + 3pi/4  =  180  + 135  =  315   =  Q4

-13pi/5  =  - 2pi - 3pi.5  =    360   - (3pi/5)(180/pi)  =   360  - 108  =  Q3

43pi/3  =  14pi + pi/3  =   Q1

-27pi/7  =  -3pi  - 6pi/7  =   180 - 6pi/7 =   180 - (6/7)(180) =  180 - 154.3  = Q1

 -5pi/4  =   -pi  - pi/4  =   180  -  45  =   135   =  Q2

11pi/6 =   pi + 5pi/6   =  180 + (5/6)(180) =  180  + 150  =  330  =  Q4

y  =(1/3)x  - 8

a.  Note  that the points  (0,-8)    and   ( 3, -7)  are on the  graph

So...these points reflected across  the x axis become    ( 0,8)  and (3,7)

And the slope of the line containing these points   is  (  7-8) / (3 - 0)  =   -1/3

So.....the  equation of the  new line  is

y = (-1/3) x +  8

Translated  up  17 units gives  us     y = (-1/3)x +  8 + 17   →  y =(-1/3)x + 25

Connect OC

Then  angle BAC   =  62°......but angle BOC  is a  central angle  intercepting the  same arc as BAC.....therefore, its measure is 2*BAC =  124°

So  in triangle  BOC.....angle  BOC  = 124° and  OB  and OC  are equal radii

Therefore  angle OBC  = angle OCB  =   (180 -124)  /  2  =   56 / 2  =    28° 

pt  = (2p - 6)(t - 1)

pt  = 2pt -2p -6t + 6

pt - 2p -6t + 6    =  0

pt - 2p =  6t - 6

p ( t - 2)  =  6t - 6

p =   (6t -6)

       ______

         t  - 2

t         p

3      12

4       9

8       7

......

Look  at  the graph here  ....  https://www.desmos.com/calculator/knrv5knjuc

Letting    t   =  x     and y  = p.....note  that p will be maximized  when  t   =  x    = 3

And when   t  = 3     then p  =12  =   the greatest number of problems you can work in one hour

sqrt (75 -27)  =   sqrt  ( 48)  = sqrt  ( 16 * 3)   = sqrt  (16) * sqrt (3)   = 4sqrt (3)

Wolfram gives the following numbers: https://www.wolframalpha.com/input/?i=+ab+%3D+2a+%2B+5b%2C+integer+solution

a = -5 and b = 1
a = 0 and b = 0
a = 3 and b = -3
a = 4 and b = -8
a = 6 and b = 12
a = 7 and b = 7
a = 10 and b = 4
a = 15 and b = 3

Okay, with your help, I understand it.  Here's what I did:   

When you fold the corner up, that leaves          BP  =  8 – 5  =  3  

And when you fold the other corner down         CQ  =  8 – 5  =  3  

So                                                                PQ  =  BC – (BP + CQ) 

                                                                    PQ  =  8 – 6  =  2  

Once you have PQ = 2 you can calcularate QR  

PQ is a diagonal of a square,               2              2 • sqrt(2)  

so the side of the square is               –––––   =   –––––––––  

                                                       sqrt(2)              2  

The 2's cancel out leaving        QR = sqrt(2) 

And PR is the same                PR = sqrt(2)  

                                                        1

Thus the area of triangle PQR is         ––– • sqrt(2) • sqrt(2)  =  1  

                                                        2    

Thanks.

Ron  

I had to edit it several times to get stuff to align properly.  What happened to WYSIWYG?  

_.

Suppose a and b are integers. How many solutions are there to the equation ab = 2a + 5b?

Looks like four to me:

Hmmm.....    but you mentioned your username in your post.....