All Answers 356020 Answers

nvm i got it it was 1

The greatest prime factor is 17.

f(2) = 1/5.

a = 19/2.

Srt the RATES equal and solve for 'x'

(x+5) / (x-1) = (x+11)/(x+2)       solve for 'x' (if possible )     Start by cross multiplying, simplify then use Quadratic FOrmula

              ..... Post if you need more help....

Then   speed in MPH  = (x+5)/(x-1)     or     (x+11)/(x+2)       <===== they should be the same for the Q

  Periods = 4 x 18 = 72

  interest per period  .05 / 4

1000 ( 1+ .05/4)⁷²      reperesents the money when she turns 18

Thank you so much!

y = k/x²

x²y= k    =  2² *10 = 40

x² y = 40

x² (4000) = 40

x² = 1/100

x = 1/10

a1 + 3d  = 8      8-3d = a1

a1 = a2 - (a2 +d)      - a1 =  d    

8 - (3*-a1) =a1

                                                 8 + 3a1=a1

                                                 8 = -2a1

                                                 a1 = -4               d = 4  

-4   0    4     8

An internet search might well produce a closed form expression for the n'th term.

You’re right; there is one:

\(L_n = \dfrac{1-sqrt(5)}{2}^n + \; \dfrac{1+sqrt(5)}{2}^n \)

The sequence starts with n=0, so the 100^th term is n=99

n=99;((1+sqrt(5))/2)^n+((1-sqrt(5))/2)^n

The web2.0calc has a rounding error greater than -2.6

Wolfram gives (489526700523968661124). Dividing this by (20) gives a remainder of (4).

GA

I decided to verify this with programming and found that my program gave the answer as 974. I didn't include details about distribution with that, but a little more poking around with it revealed that the program only found 188 instances of "threes" to your 189. I don't blame you for getting that wrong, counting is hard

Good question.   

Thanks Geno and  Ginger.

It is good to be reminded of Morgan Tud    

The queen misses those days and that scoundrel      

There might be an easier way to do this but this is what I have learnt. (I think from Heureka)  Thanks Heureka

How many integers n satisfy 0 < n < 60 and 4n=6 (mod 7)

4n=6  (mod 7)

\(4n=6\mod 7\\ 4n=7k+6\\ 4n=8k+4-k+2\\ n=(2k+1)+\frac{2-k}{4}\\ let \;\;a=\frac{2-k}{4}\\ \text{after rearranging }k=2-4a\\ \text{substituting back in}\\ n=2(2-4a)+1+a\\ n=5-7a\\ \text{n is between 0 and 60}\\ n=5,12,19,26,33,40,47,54 \)

So I found 8.

Since f(1)=32

and

f(0)=47

I think that the sum of the coefficients is  32-47 = -15

This is my logic

Since f(0)=47,    the constant must be 47    this is not a coefficient

Consider the polynomial to be

\(f(x)=ax^n+bx^{n-1}+cx^{n-2}+\dots+47\\ f(1)=a+b+c+\dots +47=32\\ so\\ a+b+c+\dots +47=32\\ a+b+c+\dots +(\text{coefficient of x)}=32-47\\\)

A short computer program gives the 100'th term as 792070839848372253127, so the remainder on dividing by 20 is 7.

(An internet search might well produce a closed form expression for the n'th term.)

Morgan Tud’s Magick of the Quadratic gives an animated formulaic and pictographic overview of the process. 

In triangle ABC, CA=4sqrt(2), CB=4sqrt(5), and A=60 degrees. What is B in degrees?

Hello Guest!

Sine law

\(\frac{sin\ \alpha}{4\sqrt{5}}=\frac{sin\ \beta}{4\sqrt{2} }\\ sin\ \beta=\frac{0.5\cdot 4\cdot \sqrt{2}}{4\cdot \sqrt{5}}=\sqrt{\frac{2}{5\cdot4}}=\sqrt{0.1}\\ \)

\(\beta=18.435°\)

  !

3.6 time 10^3 - 6.3 times 10^2 step by step

Hello Guest!

\(3.6 \times 10^3 - 6.3 \times 10^2\\ =3.6\cdot 10\cdot 10\cdot 10 -6.3\cdot 10\cdot 10\\ =3.6\cdot 1000- 6.3\cdot 100\\ =3.600\cdot 1000-6.30\cdot 100\)

    Point three places to the right

                               Point two places to the right 

 =  \(3\ 6\ \ 0\ 0-630=2970\)

  -  \(\underline{ _1\ 6_13\ 0}\)

     \(2\ 9\ \ 7\ 0\)

\(3.6 \times 10^3 - 6.3 \times 10^2=\color{blue}2970\)

  !

Thanks

If you start with the equation:  ax² + bx + c  =  0

and complete the square on this equation, you will end with the quadratic formula.

For the equation to pass through the point (0,4), the value of a must be 4.

Divide 324 by 4 to get the value of 81. This you can get by 9².

So, try the function:  y  =  4 · 9^x.

Let  c  =  -8/9,  then the  x⁴  term will disappear.

What is the radius of the circle inscribed in triangle ABC if AB = 12, AC=14, BC=18? Express your answer in simplest radical form.

Hello Guest!

The law of cosines:

\(18^2=14^2+12^2-2\cdot 14\cdot 12\cdot cos\ \alpha\\ cos\ \alpha = \frac{14^2+12^2-18^2}{2\cdot 14\cdot 12}=0.\overline{047619}\)

\(\alpha=87.271°\\ \)

\(cos\ \beta =\frac{18^2+12^2-14^2}{2\cdot 12\cdot 18}=0.\overline{629}\)

\(\beta=50.977°\)

The functions of the bisector in A and B:

\(f_A(x)=tan\ (\frac{\alpha}{2})\cdot x=0.95346\ x\)

\(f_B(x)=-tan\ (\frac{\beta}{2})\cdot x+12\ tan\ (\frac{\beta}{2}) \\ f_B(x)=-tan\ (25.4886°)\cdot x+12\ tan\ (25.4886°) \\ \color{blue}f_B(x)=-0.47673\ x+5.72078\)

Equate the functions

\(0.95346\ x=-0.47673\ x+5.72078\\ 1.43019\ x=5.72078\\ x=4\\ \color{blue}y=3.81385\)

The radius of the circle inscribed in triangle ABC is 3.81385.

  !

umm  maybe

A ray is the locus of collinear points radiating from a given start point.