All Answers 358468 Answers

Since  CD = CB.....then angles DBC  and BDC  are equal

And since angle ADC  = 116°.....then angle BDC  = 180 - 116  =  64° 

So  angle DBC  = 64°

Then angle DCB  =  180 - 2(64)  = 180 - 128   =  52°

And since  AB  = AC.....then  angles ABC and ACB  are equal

And angle ABC  = angle DBC  

So angle ACB  = 64°

And angle ACB  = angle DCB + angle ACD

So

64  =  52  + angle ACD      subtract 52 from both sides

12°  = angle  ACD

a

What we start with  -  What is cut off  =  Amt left over....so........

(49 * 6x)  -  ( 49 * 2y)  =

49 (6x - 2y)  =

49*2 (3x - y)  =

98 (3x - y)  =   amount leftover

3. 

Here's the graph :  https://www.desmos.com/calculator/zgv7tpvl0y

          x^2 - 3x                     x ( x - 3)                   x

     ___________   =     _____________  =   ______ 

     2x^2 - 3x - 9              (2x + 3) (x - 3)           2x  + 3

The vertical asynptote  is the x values that make the  denominator = 0

These are   x  = -3/2 

The  horizontal asymptote  is y  = 1/2   [the ratio of the coefficients on the variable x in the numerator/denominator ] 

The "hole"  occurs at    x - 3  =  0     ⇒   x  = 3

And the associated y value is   3 / {2(3) + 3]  =  3 / 9  = 1/3

So...the hole is at (3, 1/3)

The graph shows that the x and y intercepts occur at  (0,0)

Correct   !!!!!

2.    f(x)  =  3x^3 - 6x^2 + 3x + 6

Here's the graph :  https://www.desmos.com/calculator/68wqsc5sft

Possible positive roots  ..  f(x) has  two sign changes so   two  or zero  possible positive roots

f(-x)  = -3x^3 - 6x^2 - 3x + 6  =  1 sign change  so  one or zero posiible negative roots

Rational Roots Theorem  shows that the possible rational  roots are   ± [ 1/3, 2/3, 1, 2, 3, 6 ]

Upper Bound

2 [ 3    -6    3     6  ]              3  [  3     -6    3      6  ]

            6     0    6                                9    9    36

    ______________                ________________   

     3     0     3    12                      3     3    12    42

All the signs on the bottom row  are positive whne x  = 3.....so this is the upper bound on zeroes

Lower Bound

-1  [ 3   -6    3    6  ]

            -3    9   -12

    ______________

      3    -9   12   -6

We have alternating signs on the bottom row.....therefore....x =-1  is the lower bound on zeroes

The y intercept  is  (0,6)

The x intercept  (and the only real root)  is  x ≈ -.696

I'm just tired... 

1.   6x^2 - x <  2        rearrange as

6x^2 - x - 2 <  0      factor the left side

(2x +1) (3x - 2) < 0

We have three possible  solution intervals   (-inf, -1/2)  , (-1/3, 2/3)  and ( 2/3, infinity)

Note that if   -1/2 < x < 2/3...  this inequality will be true

So......the solution interval  is   ( -1/2, 2/3)

On a number line, draw open circles at  -1/2  and 2/3    and connect these

Subtract 2 from both sides

6x²-x-2<2-2

Simplify

6x²-x-2<0

Find the signs of the factors of 6x²-x-2

Identify the intervals that satisfy the required condition: < 0

-1/2 < x < 2/3

Hope this helps. If not, sorry mate.

now arnt we all the same

I believe it would be this,

2x² + 2x - 12

cause from what i remember from this that should be correct

(i wish for someone else to correct me if im wrong)

#whoskeepingcount

CR

ik that but i don't know the answer to 2x squared + 4 

id on't remember how to do that

bored as h***l and want to go home from high school LOL

it would seem you have to combine your xes

in this case...

2x^2 -2x +4x -12

Hope this helps

#whoskeepingcount

CR

bored as h***l and want to go home from high school LOL

okok good that's the answer i got :) 

thank you to Melody and Guest!! <3

Thanks guest

Give the first terms .. you could use either but recursive would be easier because you just have to keep multiplying by 3.

You sum is correct.

I just used the sum of a GP to solve it   

\(Sum=\frac{a(r^n-1)}{r-1}\\ Sum=\frac{1(3^{40}-1)}{3-1}\\ Sum=\frac{3^{40}-1}{2}\\ Sum\approx 6.0788\times 10^{18}\)    

omg, Same!!

(less of the school part though... more like college i would say)

well... here

-5 * -1 / -2

= -2.5 or  \(-2 {1 \over 2}\)

(dont quote me on the LeTeX first time trying it out)

#whoskeepingcount

CR

Here is my attempt:

1, 3, 9, 27, 81, 243, 729........etc.
3^0, 3^1, 3^2, 3^3, 3^4, 3^5, 3^6......etc.
On day 6, there will be:3^(6 -1) =3^5 =243 people.
Total number of people by the 6th day =S(n) =a(1) * (1 - r^n) / (1- r)  =1 * (1 - 3^6) / (1 - 3) =1 * (-728 / -2)= 364   people, or:{1 + 3 + 9 + 27 + 81 + 243} =364 people.
Or: ∑[3^(n - 1), n, 1, 6] = 364
On the 40th day =3^(40 - 1) =3^39 =4,052,555,153,018,976,267 people !!!.
Your last summation: ∑[3^(n - 1), n, 1, 40] = 6078832729528464400 - This is total number of people from day 1 through 40th day.

Cerenetie can you find a group of points really close together? Also is there a point where it is alone with no points around it? 

I assume you are just meant to answer from a visual inspection.

What do you think?

That is fine. So long as the answer is correct we can all use whatever method we want.