All Answers 358524 Answers

My friend plays Osu

Here's  part (b)

It's  the  graph of  f(2x)

Now....all you  have to  do for  part (d)  is to  shift this graph over to the  right  by  8 units  and  connect the  dots

Each  point shown will have  8  added to its x coordinate and  y will not change

Let me know if you have  trouble with this

My classroom has 11 rows of chairs, with 11 chairs in each row. The chairs in each row are numbered from 1 to 11. How many chairs have odd numbers?

We  have   11 * 11  = 121 chairs

So....the number of  "odd" ones  =  120/2 + 1 =   60 + 1  =    61

Thanks, Cal   !!!!

Those  are  going to  involve  drawing graphs......this is a little time-consuming....I'll see if I can do this in a little while, because I have other questions to answer.....sorry!!!

In right triangle  ABD, AD  = sqrt [ BD^2  - BA^2] =  sqrt [ 15^2  -  12^2]  =sqrt [ 225 -144] = sqrt [81  = 9

So...its area = (1/2)product of the legs  =(1/2) BA * AD =  (1/2)12 * 9  =  54

And  in right triangle BDC, DC = sqrt [ BC^2  - BD^2] = sqrt  [ 17^2  - 15^2] = sqrt [ 289 - 225] = sqrt [ 64] = 8

So....the  area of triangle  BDC  =(1/2) (8)(15)  = (1/2) (120) = 60

So....the  area of ABCD =   54 + 60 =   114

Also...

What about parts b and d?

(c)   f(2x - 8)    just  takes  the points that we  found in (a)  and shifts them to the  RIGHT by 8  units

So

(-2,4)   becomes   (-2 + 8, 4)  =   (6,4)    so  c   = 6

And

(2, -4) becomes  ( 2 + 8, -4)  =   (10, -4)   so d =  10

Thanks

I know  a couple of these

(a)  f(2x)    makes the  graph 1/2 as wide as the  original

So note  that   the points   (4, -4)  and  (-4, 4)   are on the original graph

So   ( a, 4)   on f(2x)  gives us  ( -2, 4)   so   a =  -2

And  (b, -4)  on f(2x)  gives  us  (2, -4)  so  b =  2

Note  that  DG is just  the  hypotenuse of a right triangle  with legs  DO  and OG

So...its length  =  sqrt [ DO^2  + OG^2 ]  =  sqrt [ 24^2  + 10^2] = sqrt  [576 + 100] = sqrt [ 676] = 26 

1.2n - 4.4 < 5.2   add 4.4 to both sides

1.2n <  9.6     divide  both sides  by  1.2

n <  8

So.....the positive integers  hat satisfy this are

1,2,3,4,5,6,7

And  their sum =   (7)(8) / 2 =  56 / 2   =  28

Sorry for messing up. I shoulda caught my mistake. But now it's fixed. Thanks in advance for answering it!!

Note  that  the  altitude of the  right  triangle on the left =  sqrt (5^2  -1^2)=  sqrt (25 -1)  = sqrt (24)

And  ?   is the hypotenuse of the right  triangle o the right  with  legs of sqrt (24)  and 5

So

? =   sqrt [ 5^2  + (sqrt (24))^2 ]   =  sqrt  [ 25  + 24 ] = sqrt [49]   = 7

Right  triangle  ACD  is a  5 -12 -13  Pythagorean Triple

So AD  = 12

Since AB  = 15.....then   BD  =sqrt  [ AB^2 - AD^2] = sqrt ( 15^2 - 12^2 ] = sqrt  [225 - 144] = sqrt [81] = 9

So....the  area  of triangle  ADB  = (1/2)(BD)(AD)  = (1/2)(9)(12)  = 54    (1)

And  the area of triangle ACD = (1/2)(5)(12) = 30    (2)

So the area of triangle ACB = (1)  - (2)  =  54  - 30  =  24

\(\frac{1}{3}<{\frac{x}{5}}<{\frac{5}{8}} \)

Take  this by parts

1/3  <  x/5                              cross-multiply           x/5   <   5/8

5 < 3x    divide  both sides  by 3                           8x  <  5 * 5

5/3 < x                                                                    8x  <   25       divide both sides  by 8                                                            

  x > 5/3                                                                    x < 25/8

So the solution is

5/3  <  x  <  25/8

I edited it. now it should be there.

-PharoaCarl

Sorry I almost forgot to put in the questions. Dumb me

-PharoaCarl