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To solve YZ, you use tangent. 

Tangent is opposite over adjacent.

Tan 28 = 26/ YZ

you can solve for YZ

What..? I dont understand..

f(x)  = (x - 3)^3  - 4x^3 (x + 2)^3 + 3(x - 1)^6      expand this

f(x)  = -x^6 - 42 x^5 - 3 x^4 - 91 x^3 + 36 x^2 + 9 x - 24

The end behavior will be determined by the first term...its end behavior "overpowers" the end behavior of the other terms

As x  approaches negative and positive infinity, f(x) will approach negative infinity

So "A"  is correct

Look at the graph here to get a feel for this :

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

Thank you so much!! I appreciate it! 

64. 

You tangent for 64.

sin 28 = 26 / YX

sin 28 is about 0.27

0.27 = 26/YX

you can solve for YX

Suppose ten DISTINCT, POSITIVE integers have a median of x. ("DISTINCT integers" means that no two integers are the same.)

What is the smallest the average of those ten integers could be?

Explain your answer in complete sentences.

hint: Read every word in this problem carefully... especially the words "DISTINCT" and "POSITIVE"!

fixed yo' question

you're welcome.

Are you missing something? Median of what?

Solve for x:
x^2 - 10 x + 36 = 0

Solve the quadratic equation by completing the square.
Subtract 36 from both sides:
x^2 - 10 x = -36

Take one half of the coefficient of x and square it, then add it to both sides.
Add 25 to both sides:
x^2 - 10 x + 25 = -11

Factor the left hand side.
Write the left-hand side as a square:
(x - 5)^2 = -11

Eliminate the exponent on the left-hand side.
Take the square root of both sides:
x - 5 = i sqrt(11) or x - 5 = -i sqrt(11)

Look at the first equation: Solve for x.
Add 5 to both sides:
x = 5 + i sqrt(11) or x - 5 = -i sqrt(11)

Look at the second equation: Solve for x.
Add 5 to both sides:

 x = 5 + i sqrt(11)          or         x = 5 - i sqrt(11)

The answer is "C"

The exact answer for RQ =8sqrt(5) =17.88854.......=17.9 - rounded to the nearest tenth, as they ask you to do.

17.9 is the correct answer

Not fully 

8*sqrt(5)=17.8885438=17.9

Ty! :D

I do not think that is the correct answer.

The two RQ are not equal.

That too lol but one of the answers to a previous question was 5/2 and the way I wrote 512 looks like 5/2

61)

16/RQ = RQ/20   The cos of RQS = cos of RQP.  Cross multiply

RQ^2 =16 x 20

RQ^2 = 320

RQ = Sqrt(320)

RQ =8sqrt(5)

Or maybe the fraction sign looks like a 1 :)

5 / 2

oh geez sorry, that's what I meant lol

SHould be 320......not 360

Ty :D

We first divide 13/5 to find the largest integer in this fraction.

13/5 = 2 R 3

The answer you seek is 2 3/5

i hope this helped,

gavin.

Omg this whole time I thought the answer said 5/2 (cuz my teacher gave me answers but I had to solve the problem to get the answer) and it says 512 I was looking at the wrong problem XDD ty

13/5 = 13 divided by 5  =  2  3/5

Need help with 58 and 61

Since the perimeter of rectangle ABCD is 60 and has a length of 20, the width is:

\((60-(20\cdot2))\div2=10\)

Since they are similar, we can set the equation:

\(\frac{20}{32}=\frac{10}{x}\)

Solving for x, we have 

x = 16

Since we know the width and the length, we know the area = 16 * 32 = 512

The answer you are looking for is 512.

I hope this helped,

gavin