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Let \(f(n)\) return the number of distinct ordered pairs of positive integers \((a, b)\) such that for each ordered pair,  \(a^2 + b^2 = n\).
Note that when \(a \neq b\), \((a, b)\) and \((b, a)\) are distinct.
What is the smallest positive integer \(n\) for which \(f(n) = 3\)?

The smallest positive integer n = 50.

\(f(50) = 3 : \\ \quad 1^2 + 7^2 = 50 \\ \quad 5^2 + 5^2 = 50 \\ \quad 7^2 + 1^2 = 50\)

the next:
\(f(200) = 3: \\ \quad 2^2 + 14^2 = 200 \\ \quad 10^2 + 10^2 = 200 \\ \quad 14^2 + 2^2 = 200 \)

the next:

\(f(338) = 3: \\ \quad 7^2 + 17^2 = 338 \\ \quad 13^2 + 13^2 = 338 \\ \quad 17^2 + 7^2 = 338 \)

the next:

\(f(450) = 3: \\ \quad 3^2 + 21^2 = 450 \\ \quad 15^2 + 15^2 = 450 \\ \quad 21^2 + 3^2 = 450\)

the next:

\(f(578) = 3: \\ \quad 7^2 + 23^2 = 578\\ \quad 17^2 + 17^2 = 578\\ \quad 23^2 + 7^2 = 578\)

the next:

\(f(800) = 3: \\ \quad 4^2 + 28^2 = 800 \\ \quad 20^2 + 20^2 = 800 \\ \quad 28^2 + 4^2 = 800\)

\(\ldots\)

The terms that will be divisible by 3  are the

2nd, 5th, 8th, 11th  etc.

So.....

2 + 3(n - 1)  ≤  100

2 + 3n - 1 ≤ 100

-1 + 3n ≤ 100

3n ≤ 101

n ≤ 101/3

n =  33  terms

I don't know how to solve this using permutations and combinations. Maybe one of the mathematicians here can do that. However, it is very easy to write a short computer code to do the counting for us:

There are 1,321 six-digit integers that sum up to 22 and they begin with:
200488  200668  200686  200848  200866  200884  202288......and end with:  882400  884002  884020  884200  886000 - for a total of 1,321 such integers.

We need to find the  x intersection points  of

x^2 + y^2   = 729     and

y = (-35/31)x + 35

So....subbing the second equation into the first for y  we have that

x^2  +  [ (-35/31)x + 35]^2  = 729     

x^2  +  1225x^2/961  -2450x/31  + 1225  = 729

[961 + 1225]x^2 / 961 - 2450x/31  + 496  = 0     multiply through by 961

2186  x^2  -  75950x  + 476656  =  0

Using the quadratic formula    we have that 

75950  ±√ [ (75950^2 - 4(2186)(476656) ]

_____________________________________

         2 *  2186

Evaluating this we get that  x =  8.221  and  x = 26.523

Putiing the first value into  the linear equation we have that

y = (-35/31)(8.221) + 35   ≈ 25.718

Putting the second vallue into the linear eqation we have that

y = (-35/31)(26.523) + 35  ≈ 5.055

So....the intersection points of the circle and line of travel  are ( 8.221, 25.718)  and (25.523, 5.055)

And  I just used the distance formula twice to solve the rest of the problem

Yeah....I can work it out for you.....in just a few....

Sorry for the bother but is there any chance you could show me how to do it algebraically?

It's review for a test and I won't have acess to a graphing calculator.

The min wil occur  when  x  = -12 / [2 * 1]  = -12.2  = -6

So.......the min for y will be

(-6)^2  + 12(-6)  + 5  =

36   -  72   + 5  =

-36  + 5  =

-31

Let the transmitter be located at (0,0)

The equation for its effective coverage area will be a circle  centered at  (0,0)  with a radius  of  27

So....the equation is   

x^2 + y^2  = 27^2

x^2 + y^2  = 729

Let the point 35  miles north of the transmitter be  (0, 35)

Let the point 31  miles east of the trasmitter   be (31,0)

The line connecting these (the line of travel)  will have a slope  of  [ 35 - 0] / [ 0 - 31]  =  -35/31

The equation of this line will be

y  = (-35/31)x  + 35

We could solve this  algebraically, but a graph seems easier to deal with  

See here : https://www.desmos.com/calculator/ksd2p5jdwl

The signal will be received  betwen the points  (8.221 , 25.718)  and (26.523, 5.055)

The distance between these points  is  

√ [ ( 26.523 - 8.221)^2  + ( 25.718 - 5.055)^2 ]  ≈  27.6 miles     (1)

The distance  between  (0, 35)  and (31,0)  =  √ [ ( 31)^2  + ( 35)^2 ]  ≈  46.8  miles  (2)

So....you will receive the signal  about  (1)/ (2)    =  27.6 / 46.8  ≈  59%  of the drive

thank you guys so so so much! <3 
i really really realllyyyyyyy appreciate it ::)

√(75) -√(12) - √(27)  =

√[25 * 3]  - √ [ 4 * 3]   -  √ [ 9 * 3 ]    =

5√3   -  2√3  -  3√3  =

0

@Melody @CPhill thanks :)

Maybe you could put all distractions aside, and make it harder for you to reach them. For example, you could make a study place/table somewhere that doesn't include distractions, such as a computer, phone, game, etc (unless you need some of those things to look something up, or if your resources to study are online). You could also try joining or creating a study group and study with them to make you feel more motivated, and taking breaks if you need them --just be sure to go back to studying in about 10-15 mins! Also, if you procrastinate, even if it's as short as let's say an hour, the task you are trying to get done will seem bigger and bigger, and you're likely not to accomplish as much as you could've.

Hope this helps !

I can usually avoid studying in most classes by paying attention and doing hw and stuff.

If that fails, then if the teacher gives you some list of things to know for the test thats usually a good place to start. I usually look over that and fill in stuff I don't know from notes and stuff. (this is also a great resourse for cramming right before the test, totally don't know that from experience)

When I need to learn a list of vocab words I use Quizlet. Also comming up with weird associations between words and definitions helps me remember things, like what a word sounds like or how its spelled. Even sitting there trying to come up with something to remember a word by helps.

It usually helps me to look at something, come up with a few words or short phrases about it, say it in my head/wisper it, say it w/o the paper a few times, then come back later and see if I can still remember it. This is good for knowing a concept or comparing/contrasting stuff, where you really only need to know the general idea, then be able to lengthen and add detail to it on the test.

Anyways my study habbits are atrocious, so take this with a grain of salt (or maybe a whole shaker full).

Slayer, you were SO CLOSE ... you missed the sign in the last line. 

–4 * 6  =  –24 

.

-16^2 +32t +15

The max height occurs when t  =   (-32) / (2 * -16)  =   -32/-32  =   1

So.....the max height is

-16(1)^2  + 32(1)  + 15 =

-16  +  32  + 15  =

31 ft

Problem 3

They are similar because   the missing angle in the first triangle is 180 - 123 - 36  = 21°

So....two  angles of  triangle CDE  = two angles of triangle CAB

So....by  Angle-Angle congruency, the triangles are similar

We have 12 pens in total

P(first three pens are blue with replacement)  =  3/12 * 3/12 * 3/12  =

(1/4) * (1/4) * (1/4)   =

1/ 64

The minimum occurs  when b  =   -5 / (2 *1/2) = -5 / 1    =    -5

Thanks Ginger.

That was a joy to read.

Did 315 graduate?  I guess he must have. I have not heard from him for ever. It was nice to be reminded.

He was one of our many very pleasant members. 

When people do a job for a long time they become unappreciated. Or at least the appreciation that is held for them is less often voiced.

This is even more true when people give their time and expertise at no charge.

Rom,  more people here appreciate your efforts than you realize.  

I am one of them.  I am sorry that I do not thank people like yourself often or well  enough.

Whether you chose to return or not, I just want you to know that. 

All circles are similar BECAUSE the ratio of the circumference to the diameter is ALWAYS the same for all circles, that is: Circumference / Diameter = Pi

ah, i see :) 
thanks

4) All circles are similar because they all lack angle measurements. Like triangles have angle measurements, and squares and other shapes. Except circles. That's one thing that makes them all similar.

Your answer is 0.

sqrt(75)-sqrt(12)-sqrt(27) = 0

-22 + 18 = -4

-4 * 6 = 24

24's your answer

The slope is -4/2 = -2.

The y-intercept is (0, -2)

We can use this information to write an equation for the function and find the x-intercept, the x-intercept is the value of x when f(x) = 0:

f(x) = mx + b

f(x) = -2x -2

0 = -2x - 2

2 = -2x

-1 = x

The x-intercept is (-1,0)

Hope this helps!

SS