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1     solve     6 = -4.9t^2+14t-0.4          using quadratic equation      a = -4.9    b = 14    c = - 6.4  

                                                        you will get two values of t   the time above 6 is between these times

3  Max will occur at the point between the two zeroes    -7  and 2      middle is - 2.5

      use this value for x in the equation to find the max value of the equation

The minimum value of r^2 + s^2 + is 8.

Using a computer program, a_{50} leaves a remainder of 1, and b_{50} leaves a remainder of 2, so a_{50} + b_{50} leaves a remainder of 3.

Exactly how many of the TVs are defective?

Four

Find all solutions to the system
a + b = 10
a^3 + b^3 = 812

Hello Guest!

\(a + b = 10\\ a^3 + b^3 = 812 \)

\(b=10-a\\ a^3+(10-a)(100-20a+a^2)=812\\ a^3+1000-200a+10a^2-100a+20a^2-a^3\\\)

\(30a^2-300a+1000=0\\ 3a^2-30a+100=0\\ a^2-10a+\frac{100}{3}=0\\ a=5\pm \sqrt{25-\frac{100}{3}}\\ a=5\pm i\sqrt{\frac{100}{3}-25}\)

\(a_1=5+i\cdot \frac{5\sqrt{3}}{3}\\ \color{blue}a_1=5+2.88675\ i\\ a_2=5-i\cdot \frac{5\sqrt{3}}{3}\\ \color{blue}a_2=5-2.88675\ i\)

\(b_1=10-(5+2.88675i)\\ \color{blue}b_1=5-2.88675\ i\\ b_2=10-(5-2.88675i)\\ \color{blue}b_1=5+2.88675\ i\\ \)

  !

The quadratic x^2 +5x + c has roots in the form of x = (-5 +/- sqrt(2c + 3))/2.  What is the value of c?

\(ax^2+bx+c = x^2 +5x + c\)      so what do you know so far.

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a} = {-5 \pm \sqrt{2c+3} \over 2}\)

Do some substitutions and finish it.

\(n^3 = n + 3n(n - 1) + 6 \binom{n}{3}\)

Prove by induction for n>=3

Step 1

Prove true for n=3

LHS =27

RHS=27

So the statement is true of n=3

Step 2

Assume true for n =k where k is some integer greater or equal to 3

So this statement is true:

\(k^3=k+3k(k-1)+6\binom{k}{3}\\ k^3=k+3k^2-3k+6*\frac{k!}{(k-3)!3!}\\ k^3=3k^2-2k+\frac{k!}{(k-3)!}\\ k^3-3k^2+2k=\frac{k!}{(k-3)!}\\ \)

NOW prove that it will also be true for n=k+1

i.e.   Prove

\((k+1)^3=(k+1)+3(k+1)(k)+6\binom{k+1}{3}\\~\\ LHS=x^3+3k^2+3k+1\\~\\ RHS=k+1+3k^2+3k+6*\frac{(k+1)!}{3!(k-2)!}\\ RHS=3k^2+4k+1+\frac{(k+1)!}{(k-2)!}\\ RHS=3k^2+4k+1+\frac{k!(k+1)}{(k-3)!(k-2)}\\ RHS=3k^2+4k+1+\frac{k+1}{k-2}*\frac{k!}{(k-3)!}\\ sub\\ RHS=3k^2+4k+1+\frac{k+1}{k-2}*(k^3-3k^2+2k)\\ RHS=3k^2+4k+1+\frac{k+1}{k-2}*k(k^2-3k+2)\\ RHS=3k^2+4k+1+\frac{k+1}{k-2}*k(k-2)(k-1)\\ RHS=3k^2+4k+1+(k+1)*k(k-1)\\ RHS=3k^2+4k+1+k^3-k\\ RHS=k^3+3k^2+3k+1\\ RHS=LHS\)

So if it is true for n=k, it will be true for n=k+1

Step 3

The statement is true for n=3 so it is true for n=4, n=5, .....

Hence the statement is true for all integer values of n greater or equal to 3.

LaTex coding:

k^3=k+3k(k-1)+6\binom{k}{3}\\
k^3=k+3k^2-3k+6*\frac{k!}{(k-3)!3!}\\
k^3=3k^2-2k+\frac{k!}{(k-3)!}\\
k^3-3k^2+2k=\frac{k!}{(k-3)!}\\

(k+1)^3=(k+1)+3(k+1)(k)+6\binom{k+1}{3}\\~\\
LHS=x^3+3k^2+3k+1\\~\\
RHS=k+1+3k^2+3k+6*\frac{(k+1)!}{3!(k-2)!}\\
RHS=3k^2+4k+1+\frac{(k+1)!}{(k-2)!}\\
RHS=3k^2+4k+1+\frac{k!(k+1)}{(k-3)!(k-2)}\\
RHS=3k^2+4k+1+\frac{k+1}{k-2}*\frac{k!}{(k-3)!}\\
sub\\
RHS=3k^2+4k+1+\frac{k+1}{k-2}*(k^3-3k^2+2k)\\
RHS=3k^2+4k+1+\frac{k+1}{k-2}*k(k^2-3k+2)\\
RHS=3k^2+4k+1+\frac{k+1}{k-2}*k(k-2)(k-1)\\
RHS=3k^2+4k+1+(k+1)*k(k-1)\\
RHS=3k^2+4k+1+k^3-k\\
RHS=k^3+3k^2+3k+1\\
RHS=LHS

1 is a positive integer and if n=1 then the expression makes no sense.

So I suppose that is an instant disproof.

If it is true at all, n would have to be a positive integer greater or equal to 3

I have absolutely no idea what the last sentence in the question is referring to.

7 * 5!   sounds good to me.

Put one in the middle, 7 possibilities.

Take one at random and put it in a position.  This marks the ästart of the outer ring.

Then there are 5 left so they can be ordered in 5! ways

These 3 triangles are similar. 

All three triangles are 30-60-90 which means that the length of a short leg is half of a hypotenuse.

ED = 36 / 2³ = 4.5

CD = 4.5 * tan(60) = 7.794228634

Thanks guest.

I suppose I was just making too big a deal out of it.

I was looking for a general solution, ... the question is plural.

Between 100 and 999, there are:999 - 100 + 1 =900 three-digit integers.

900 / 13 = 69 of them are multiples of 13

143  273  403  533  663  793  923  Total =  7 of those 69 are multiples of 13 and end in 3.

Probability is = 7 / 900 - that it is multiple of 13 and ends in 3

H is the orthocenter of acute triangle ABC and the extensions of AH, BH, and CH intersect the circumcircle of traingle ABC at A prime, B prime and C prime. We know angle AHB : angle BHC : angle CHA = 2 : 5 : 8. Find angle AprimeBprimeCprime in degrees.

angle AHB : angle BHC : angle CHA = 2 : 5 : 8       (48º,120º,192º)

This ratio is NOT possible!!!   These angles cannot be larger than 180º.

In what bases, b, does (b+7) diveide into (9b+6) without any remainder?  

Maybe I'm not understanding the question, because I found it pretty easy to determine that  

                             (9)(12) + 6            114  

                             —————    =    ———    =    6     

                               (12 + 7)                19  

_.

I would like to see an answer to this one too 

GF = (24 * 8) / 23        you can do the rest

Let's call (a,b,c) a smiley face if b is less than a and b is less than c, because when we plot the graph, we get a happy face!  And if b is greater than a and b is greater than c, that's a frowny face, because we get a frowny face when we turn a smiely face up-side-down.

There are other kinds of faces like smirks (like a is less than b and b is less than c) and neutral faces (like when a is equal to b and b is equal to c).  If the face is neutral, then a equals b and b equals c, so when we choose a, b, and c are also chosen, and there are n choices for a, so there are n neutral faces.

Now we count the number of smirks.  There are n ways to choose a, and there are n - 1 ways to choose b.  We also multiply by 3, because the value that we chose for a could have also been the value of b, or the value of c.  So there are 3n(n - 1) smirks.

Now we count the number of smiley faces.  There are n ways to choose a, then n - 1 ways to choose b, then n - 2 ways to choose c.  So there are n(n - 1)(n - 2) = 3C(n,3) smiley faces.  By symmetry, there are 3C(n,3) frowny faces.

Therefore, the total number of faces is n^3 = n + 3n(n - 1) + 6C(n,3).

angle BDA = arctan(3/2) = 62.567 degrees

because ABC is an issosolies right triangle, the angles measure 90, 45 and 45 degrees

the measure of ABD=45 degrees and side AC is trisected --> the measures of ABE EBD and DBC are all 15

in triangle DEB we only know angle DBC BUT we can figure out CDB

180=DCB+BCD+CDB

180=15+45+CDB

CDB=120

CDB+BDE=180 because they make a line

120+BDE=180

the measure of BDE=60 degrees

yes, maybe :)

People learn through their mistakes.  It is no big deal.

Statements (3) and (4) must be true.