All Answers 357301 Answers

SECOND WAY (probably easier for you)

Use you calculator to get values for     cot(0.5*3pi),    cot(0.5*0),    cot (0.5*2pi),  Cot(0.5*-2pi)   and cot(0.5*pi/2)

If you get a an error then that vaue of x is an asymptote.

If your calc does not have cot then use   1/tan( )

Lets look at this anyway. You should not rely on desmos, that is just for checking.

FIRST WAY  (The most formal way)

$y=3cot(0.5x)-4$

The 4 only pushed the graph up or down so it has nothing to do with aspymptotes.

The 3 just exaggerates the graph in a up down direction so that has nothing to do with it either.

so that asymptotes will be the same as for

$y=cot(0.5x)$

now

$y=cot(0.5x)=\frac{cos(0.5x)}{sin(0.5x)}\\ \text{You cannot divide by 0 so the asymptotes will be where sin(0.5x)=0}\\ 0.5x\ne n\pi \qquad where \;n\;is\;an\;integer\\ so\\ x\ne 2n\pi$

so x canot be 0 or  pi or -pi    they are three of the asymptotes.

a(b)^(cx) = d, solve for x
Take the log of both sides:
cx log(a(b)) = log (d)
Divide both sides by: c log(a(b))
x = log (d) / c log(a(b))

Oh okay, so I was correct at first, thank you! 

OK :) 

It is the second and the third one.

The first and the forth BOTH cross the graph, so they are not asymptotes.

Okay, I guess I'm viewing something wrong, could you tell me what exactly you think the answer is?

I did check on desmos, now I think it's the second, third, and fourth option, do you agree?

Cupcake, put your graph into the desmos program like I did with theother one and then you can see for yourself.

$\sqrt{49-20\sqrt6}\\ =\sqrt{25+24-20\sqrt6}\\ \text{This might be helpful because 25 is a square and 24 is a multiple of 6}\\ =\sqrt{25-20\sqrt6+4*6}\\ =\sqrt{5^2-20\sqrt6+(2\sqrt6)^2}\\ =\sqrt{5^2-2*5*2\sqrt6+(2\sqrt6)^2}\\ =\sqrt{(5-2\sqrt6)^2}\\ =|5-2\sqrt6|\\ =5-2\sqrt6\\$

Now maybe you can attempt the second one.   

Start by factoring out the 2.

Whoa there , pardner.....evaporating ether is ENDOTHERMIC.....it must absorb heat from surrounding to go from liquid to a gas....    Being flammable has nothing to do with it....if it is BURNING...THAT would be exothermic.

A. Put ZERO in for t and calculate

B  exponential growth

C We are currently 19 years past 2000....put 19 in for t and calculate....answer will be in MILLIONS....so multiply by 1000000 to get population count. (Same process for A part)

I have not worked through this qustion but I'd start with the generic polynomial

$f(x)=ax^3+bx^2+cx+d$

Then start substituting so see if you can get any of those coefficients.

Starrt with  f(0)=0

You will probably end up with a couple of equations that you will need to solve simultaneously.

See how you go and post what you find for yourself.

If you then need more help then continue this thread.

Please no one else answer until asker guest has returned with some input of his/her own.

Yes that is correct :)

Use the the desmos online graph to help you with questions like this, or to check your answers. :)

Thank you. It's false

I've graphed it here

If it is correct then no point on  the orange lines would not be included in the trig graph.

So is it correct?

Applesoup,

Did you become a member just so that you could take points off CPhill?

He works hard for those points !

Fomula for finding the $n^{th}$ term of arithmetic sequence: $a_n=a_1+(n−1)d$ ($a_1$ is the first term, $n$ is the term you want to find (eg for finding the $7^{th}$ term, $n$ would be $7$), and $d$ is the common difference).

Formula for finding the sum of an arithmetic sequence: $S_n=\dfrac{n(a_1 + a_n)}{2}$ ($a_1$ is the first term, $n$ is the number of terms, and $a_n$ is the last term).

1)

So we find the last term with the first formula:

$10 + (8-1)2$ $=24$.

And the sum with the second formula:

$\dfrac{8(10+24)}{2}=136$

So the number of miles Mercedes will ride over the course of $8$ weeks is $\boxed{136}$

---

2)

Likewise we can solve this one with the two formulas.

We find the $7^{th}$ term with the first formula:

$15+(7-1)3$ $= 33$

And the sum with the second formula:

$\dfrac{7(15+33)}{2}=168$

So the total number of logs in the stack is $\boxed{168}$

$Q.E.D$

Row 0 has one number

Row 1 has two numbers

.

.

.

and so on

Therefore, row 42 must have 43 numbers.

The first number starts $\binom{n}{0}$with, the second number,  $\binom{n}{1}$ , and so on..

Thus, the answer is $\binom{42}{1}=\boxed{42}.$

Hope that helped!

This is why I had diffuclty specfically with A since it is a vapor but from your explanation that makes sense as to why it would release heat. Thank for your help I will also make note of this.

Well, the median is half the hypotenuse in a right triangle. And, just divide that length by 2.

Yes they can get really confusing! Is there anyway you can check the second one as well , that is the one i was having the most difficulty on?