All Answers 356000 Answers

the smallest odd number greater than 7x − 2     will always equal 7x-1

he largest odd number less than 3x + 5                is 3x+3

(7x-1) - (3x+3) = 4x - 4         a = 4 b = - 4         a+ b = 0

Since  the regular hexagon is divided into 6 congruent triangles, it simply means that each triangle is an "equilateral" triangle, as you state in the question, or all 3 of its side are equal to each other. Since the perimeter of one of these triangles is 39 inches, then each of its sides =39 / 3 = 13 inches. This is also one side of the hexagon. Since a regular hexagon has 6 sides, then its perimeter =6 x 13 = 78 inches.

You can do these problems very easily yourself,   here is screenshot of your problem solved by Web2 calculator: (even shows your required fraction)

first divide opposite by the hypotenuse or 20 / 25

then multiply the answer by sine^-1

so your answer is D:

20/25 = 0.8

sine^-1 (0.8) = 53 and some change

Well  .....just convert it.....you do know how to do that, yah?

-725/1000

-29/40

I need it in a fraction though.

-0.725    from pasting into the calculator on this site.

The function can take on positive real value, so the range is (0,infinity).

[(x-2) + (x) + (x+2)  ]  =  1/7(x-2)(x)(x+2)

3x   = x^3-4x

21 = x^2-4

0 =x^2 -25    x = -5 (throw out) or 5

so the numbers are   5-2 = 3   5   and 5+2 = 7         (as guest posted)

3 miles 1 hour

12 miles x hour

x = (12×1)/3 = 4

it's D

3 + 5  +  7= 15

3 x 5 x 7  =105

105 / 15  = 7  and 5 is the answer.

WTF! This looks like a bloody inquisition. 

You plop in here as an imperious, fucking know-it-all –and you are anonymous because ...well the AssHoleMan doesn’t want to present any kind of identity. But of course, we know who you are by your erudite syntax and presentation of pure condescending BULLSHIT!

(Ahem) What do you think about the answerers who don't bother to check their answers or read the question correctly? Don't you think that it can be irresponsible towards the users that come here to learn?

This line is particularly condescending:

Especially when this site hosts many knowledgeable mathematicians and has earned a grear reputation?

You can take your patronizing superiority and shove it up your ass!

If you were truly knowledgeable about this site, you’d know there are very few knowledgeable mathematicians on this forum.  To you, anyone who knows the multiplication tables is a knowledgeable mathematician, so it’s just a matter of perspective. In any case, you are not qualified to make the statement. 

I’ve seen your work on here, and it’s clear you are a fucking idiot. You think of yourself as a knowledgeable mathematician and educator because you’ve provided solutions to a few junior high school math questions. I’m sure you’ll start having wet dreams when you remember how to solve quadric equations. I am personally overwhelmed by your majestic presence. You belong in the forums “BB” class. BB stands for Blarney Bag and it includes Bullshit Bugs and Blarney Bankers.  This is the league of Extraordinary Batshit Stupid Gentlemen: those who’ve retired with a pension, hardened arteries, atrophied brains, and an extreme desire prove their virile intellect via mathematics.

I’m sure you belong in this class. Let me guess, you are a recently retired what ... a salesman ...insurance maybe.  Whatever... You’ve not practiced any math for forty years or more, and now that you have time, and are bored-out-of-your-gourd, you’ve decided you want to help the children of the current generation.

How fucking noble! We are soooo honored to have you. To prove it, we will shit marble every time you post.

GA

I want to politely tell you, Mr. Guest, that many of my friends at my school would have already told you to "---- off". We're not your personal slaves, we can do whatever we want. We don't have to answer your questions, and YOU need to learn some discipline. I expect you are some spoiled rich kid who gets whatever they want. Well then, welcome to the real world, where no one gives a c**p about what you think. 

It appears that the right answer should be =10,001

gcd(60786078, 25632563, 12341234, 41874187, 52975297, 27542754, 56785678, 27512751......etc.) =10,001

When you factor all of them, they all seem to share the same 2 factors: 73 x 137 =10,001.

There is probably some good technique to use.

I just found it by playing with the numbers .

One of the numbers is 3*3*5 = 45

The lowest common multiple has to be a multiple of 45         that is   3*3*5*something

when LCM is divided by the greatest common divisor then the answer is 33

33=3*11

So the other number will have to be a multiple of 11  

\(\frac{3*3*5*11*?}{\text{some factor of 45}}=33\\ \frac{(3*5)*3*11*1}{3*5}=33\\\)

So if the greatest common divisor is 3*5

The given number is    3*3*5

The smallest value of the other number will be  3*5*11 = 165

I think I have already told you what I think.

If askers want to question an answer they are encouraged to do so.

What do you think about the answerers who don't bother to check their answers or read the question correctly? Don't you think that it can be irresponsible towards the users that come here to learn? Especially when this site hosts many knowledgeable mathematicians and has earned a grear reputation?

Maybe so, hence he/she is not interested in what is presented here.

People can give wonderful answers and many/most askers do not even have the good grace to say thankyou.

Maybe they have not even seen the answer. The assignment was due an hour ago, they no longer have any interest.

Maybe the question asker solved the question using a credible source? Maybe he already solved the question himself?

No, I would not go that far.

But it is still up to the asker to decide whether or not to accept an answer.

The asker can also interact with the answerer and if they are not happy they can even request a specific answerer to intervene.

This is easier for members, they can do it by private message if they want.

They may not get the help from the specific person that they want but often they do.

Are you the person who asked this question?

If you are then why are you not drawing attention to your specific concerns.

and

If you are not then that is a part of my point. Questions are dumped here. Even the asker usually does not care if it is answered or if it is right and they are very often not interested in learning anything from the answer. If they want to learn, and they do not understand,  then they will interact with the answerers.  Errors will be corrected or at least identified.

Find the area enclosed by the graph of the parametric equations \(\begin{align*} x &= 6 \cos t \sin t, \\ y &= 6 \cos^2 t. \end{align*}\)

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{6 \cos t \sin t} \\ x &=& 3\cdot 2 \cos t \sin t \quad & | \quad \small{ 2\cos(t)\sin(t)=\sin(2t)} \\ \mathbf{x} &=& \mathbf{3\sin(2t)} \\ \hline \end{array} \)

\(\begin{array}{|rclrcl|} \hline \mathbf{y} &=& \mathbf{6 \cos^2(t) } \\\\ && & \cos(2t) &=& \cos^2(t)-\sin^2(t) \\ && & &=& \cos^2(t)- \left(1-\cos^2(t) \right) \\ && & &=& \cos^2(t)- 1 +\cos^2(t) \\ && & &=& 2\cos^2(t)- 1 \\ &&& 2\cos^2(t) &=& 1+\cos(2t) \\ &&& \mathbf{ \cos^2(t)} &=& \mathbf{\dfrac{1+\cos(2t)}{2}} \\ \\ y &=& 6 \left( \dfrac{1+\cos(2t)}{2} \right) \\ y &=& 3 \Big( 1+\cos(2t) \Big) \\ \mathbf{y} &=& \mathbf{3+3 \cos (2t) } \\ \hline \end{array} \)

\(\begin{align*} x &= 3\sin(2t), \\ y &= 3+3 \cos (2t) \end{align*}\)  this is a circle with center \((0,3)\) and radius \(= 3 \)

The area is  \(\pi r^2 = 3^2\pi = \mathbf{9 \pi} \)

Do yoy think it's a good idea to give some askers wrong answers deliberately?

The answers are not examined in detail, there is not enough time to examine every question.

It is up to members and guests to indicate, politely and with a reason, why the answer is inadequate or incorrect.

I have no problem with some answers being wrong. The asker should have enough knowledge to pick up careless mistakes and to question illogical answers.

If they do not understand they should ask questions and interact with the answerers.

Being ticked off does not prevent further posts on the thread.

If they copy the answer down without understanding, and hand it in as homework then it is good that it is wrong.  They should not be copying other people's answers anyway.

Find the sum of all possible values of the constant k  such that the graph of the parametric equations
\(\begin{align*} x &= 2+ 4\cos s,\\ y &= k-4\sin s, \end{align*}\)
intersects the graph of the parametric equations
\(\begin{align*} x&=1+\cos t,\\ y&=-3+\sin t \end{align*}\)
at only one point.

\(\text{Let the center of circle 1, $~ \mathbf{c_1} = \dbinom{2}{k} $} \\ \text{Let the radius of circle 1, $~ \mathbf{r_1} = 4 $} \\ \text{Let the center of circle 2, $~ \mathbf{c_2} = \dbinom{1}{-3} $} \\ \text{Let the radius of circle 2, $~ \mathbf{r_2} = 1 $} \\ \text{$\mathbf{\overline{c_1c_2}}$ is the distance between $c_1$ and $c_2$ } \)

There is only one point, if \(\mathbf{|\overline{c_1c_2}| = r_1+r_2 }\)

\(\begin{array}{|rclcl|} \hline |\overline{c_1c_2}| &=& r_1+r_2 \\ |\sqrt{(2-1)^2+(k-(-3))^2} | &=& 4+1 \\ |\sqrt{1+(k+3)^2} | &=& 5 \\ 1+(k+3)^2 &=& 25 \\ (k+3)^2 &=& 24 \\ k+3 &=& \pm\sqrt{24} \\ \mathbf{k } &=& \mathbf{-3 \pm\sqrt{24}} & \text{or}& k=-3\pm 2\sqrt{6} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline k_1+k_2 &=& -3 +\sqrt{24}-3 -\sqrt{24} \\ k_1+k_2 &=& -6 \\ \hline \end{array} \)

The sum of all possible values of the constant k is -6

Are incorrect answers considered adequate?