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Your last one is CORRECT!!!!

Third one

(x + 2)^2           ( y + 8)^2

_______    +    _________  =     1

     16                     81   

Center   (-2, -8)

The major  axis  =   2sqrt (81)  =  2 * 9  =   18  units long....so....the enpoints of the major axis are  18/2 =  9 units from the center

The minor  axis =  2sqrt(16)  = 2 * 4  = 8  units long....so....the endpoints of the minor axis  are  8/2  = 4 units from the center

The endpoints  are  ( -2  + 4 , -8)  ( -2 - 4, -8)   ( - 2,  -8 + 9)    (-2, -8 - 9)   =

(2, -8)   ( -6, -8)  (-2, 1)   (-2 - 17) 

Connect these with a smooth curve to graph the ellipse....here it is  :

https://www.desmos.com/calculator/jwlnfr0asu

Second  one.....the easiest way to  see  this one is to note what happens to the lead term of each polynomial  as we approach  either a large negative  or a large positive

1.    As  - 2x^3    approaches  some large negative value.....the limit of the polynomial  approaches  positive infinity.....so.....True is  correct!!!

2.  As 2x^4  approcahes some large  positive value......the polynomial  approaches positive infinity......so....True is correct

3.   As  -9x^5  approaches some  large positive value, the polynomial approaches negative infinity....so.... True is correct

ohh okay i see my mistake thank you!!

First one

Center  =  (1,4)  = (h, k)

Since the transverse axis is parallelto the y axis, the hyperbola opens up and down

The form is

(y - k)^2            (x  -h)^2

______    -     _________   =    1

    a^2                  b^2

a = 20/2  =  10  ⇒  a^2  = 100

b = 16/2 =  8  ⇒ b^2  = 64

So.....the equation is

(y - 4)^2        (x - 1)^2

______  -     _______   =     1

 100                64

Here's the graph :

https://www.desmos.com/calculator/ppuyhib2r6

Oh so my speculation was right

Thanks!

Nice, heureka....!!!!

-9√ [ 2x + 1]  >  -18       

Divide  both sides by  -9......since we are dividing an inequality by a negative, we need to reverse the inequality sign.....so....we have

√[2x + 1 ]  <  2          square both sides

2x + 1  < 4           subtract 1 from both sides

2x  <  3                 divide both sides by  2

x < 3/2

But also remember that  the result of  the expression ubder the radical  must  be  ≥ 0

So

2x  + 1  ≥  0

2x ≥  -1

x ≥  -1/2

So.....the solution  is

-1/2  ≤ x  <  3/2

We  wll  have a   "closed" circle on the left side and an "open" circle on the right  - so......the last choice

Does this make sense, Jenny  ????

hint:

the first few prime odd numbers are 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79.

we also know that the ratio of a to b to c is 3:4:5.

c also have to be greater than 5.

prime odd integers whose sum is a perfect cube (1,8,27,64,125,216,etc)

Why do you need an answer so quickly?  Are you just trying to get us to do your homework for you?

is this homework?

How...or what have you attempted to do to answer this long question?

6/x  + 2x²      Put in 3   for x

6/3  + 2 (3^2)

2 + 2(9) =    I think you can do it from here.....

There is no 'solving' to be done here....your mathematical equation is incorrect

-6 +-7 = -13

I agree with Electric because -6+-7= -6-7, which is -13.

idk but why didn't the skeleton go to the party

Why didnt the skeleton go to the party?

Could also do it like this:

Let a=log₃x.     b=log₉x.      c=log₈₁x

Then x=3^a    x=9^b=3^2b   x=81^c=3^4c

So a=2b and a=4c

But a+b+c=7 so a + a/2 + a/4 = 7 or 7a/4 = 7 or a = 4

Hence x = 3⁴ = 81

There is no 'solving' to be done here....your mathematical equation is incorrect

-6 +-7 = -13

There are two integers n that work.

The inequality is x - 4y + 5 > 0.

This works out to 216.

There are 434 possible locations for the centroid.

  a Little more detail ....how to SOLVE algebraically:  (using base change log rule)

log₁₀ x / log10 3     log₁₀ x / log₁₀ 9  + log₁₀ x / log₁₀ 81 = 7

2.0959  log x      +   1.04795 log x  +  .5239 log x  = 7

3.66783 log x = 7

log x = 1.9084

x = 81

As someone said before, by inspection, x = 81

log(81,3)+log(81,9)+log(81,81) = 7