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You are just meant to use the Pythagorean theorem.

Is that enough info?  

What does Pythagoras's theorem say?

5 numbers with a range of 5 a mean of 4 and a median of 3

How about:  3 3 3 3 8 

By positive...

In some places 0 is considered positive and negative, some neither.

I'll go with the assumption 0 is neither since where I live that's the definition.

Thus we have the pairs...

 1 26

2 25

3 24

...

26 1

1 through 26 is 26, so we have 26 pairs. There are 13 distinct pairs and times 2 for the ability to flip over. This is because 27 is odd and 27/2 isn't an integer.

If the number was odd however we would have a pair (even num.)/2 and we can't flip that.

If you don't understand anything feel free to ask!

(sorry for the late answer!)

You misunderstood the question! There are: 5! / 2 = 60 permutations. Since there are 5 letters and 2 of them are duplicates, therefore there would be: 60 / 5 =12 x 2 letters that are duplicates = 24 times that each permutation will begin with a letter "L".

In an isosceles triangle, the sum of its base and its height is 10,
and the radius of its circumscribed circle is 25/8.
Find the sides of the triangle.

\(\begin{array}{|lrcll|} \hline & c+h &=& 10 \\ \text{or} & \mathbf{h} &=& \mathbf{10-c} \\ \hline \end{array} \begin{array}{|rcll|} \hline \left(\dfrac{c}{2}\right)^2+h^2 &=& a^2 \\ \left(\dfrac{c}{2}\right)^2+(10-c)^2 &=& a^2 \\ \dfrac{c^2}{4}+(10-c)^2 &=& a^2 \\ \mathbf{a^2} &=& \mathbf{\dfrac{c^2}{4}+(10-c)^2} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline A &=& \dfrac{ch}{2} \\\\ \mathbf{A} &=& \mathbf{\dfrac{c(10-c)}{2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{r} &=& \mathbf{\dfrac{abc}{4A}} \quad | \quad b=a \\\\ r &=& \dfrac{ca^2}{4A} \\\\ r &=& \dfrac{c* \left(\dfrac{c^2}{4}+(10-c)^2 \right)}{4A} \\\\ r &=& \dfrac{c* \left(\dfrac{c^2}{4}+(10-c)^2 \right)}{4*\dfrac{c(10-c)}{2}} \\\\ r &=& \dfrac{c* \left(\dfrac{c^2}{4}+(10-c)^2 \right)}{2c(10-c)} \\\\ r &=& \dfrac{ \dfrac{c^2}{4}+(10-c)^2 }{2(10-c)} \\\\ 2(10-c)r &=& \dfrac{c^2}{4}+(10-c)^2 \\\\ (10-c)^2 - 2r(10-c) + \dfrac{c^2}{4} &=& 0 \\\\ 100-20c+c^2 -20r + 2rc - \dfrac{c^2}{4} &=& 0 \\ \ldots \\ \mathbf{\dfrac{5}{4}c^2-c(20-2r) + 100-20r} &=& \mathbf{0} \\\\ c &=& \dfrac{20-2r \pm \sqrt{(20-2r)^2-4*\dfrac{5}{4}*(100-20r)} }{ 2*\dfrac{5}{4}} \\\\ c &=& \dfrac{20-2r \pm \sqrt{(20-2r)^2- 5(100-20r)} }{ 2*\dfrac{5}{4}} \\ && \ldots \\\\ c &=& \dfrac{20-2r \pm \sqrt{4(r^2+5r-25)} }{ 2*\dfrac{5}{4}} \\\\ c &=& \dfrac{20-2r \pm 2\sqrt{ r^2+5r-25 } }{ 2*\dfrac{5}{4}} \\\\ c &=& \dfrac{10-r \pm \sqrt{ r^2+5r-25 } }{ \dfrac{5}{4}}\quad | \quad r=\dfrac{25}{8} \\\\ c &=& \dfrac{6.875 \pm \sqrt{ 0.390625 } }{ 1.25} \\\\ c &=& \dfrac{6.875 \pm 0.625 }{ 1.25} \\\\ c_1 = \dfrac{7.5}{1.25} && c_2 = \dfrac{6.25}{1.25} \\\\ \mathbf{c_1=6} && \mathbf{c_2=5} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{a^2} &=& \mathbf{\dfrac{c^2}{4}+(10-c)^2} \\ \hline \mathbf{a_1^2} &=& \mathbf{\dfrac{6^2}{4}+(10-6)^2} \quad | \quad c_1=6 \\ a_1^2 &=& \dfrac{36}{4}+4^2 \\ a_1^2 &=& 9+16 \\ a_1^2 &=& 25 \\ \mathbf{a_1} &=& \mathbf{5} \\ \hline \mathbf{a_2^2} &=& \mathbf{\dfrac{5^2}{4}+(10-5)^2} \quad | \quad c_2=5 \\ a_2^2 &=& \dfrac{25}{4}+5^2 \\ a_2^2 &=& \dfrac{25*5}{4} \\ \mathbf{a_2} &=& \mathbf{\dfrac{5}{2}\sqrt{5}} \\ \hline \end{array} \)

You misunderstood the question! It is asking for the "sum of the digits divisible by 2"! Is 1000,000,000 divisible by 2. No!. Its digits sum up to 1.

The first number whose sum of the digits is divisible by 2 =1,000,000,001 = 2 which is divisible by 2

The last number whose sum of the digits is divisible by 2 = 9,999,999,999 = 90 which is divisible by 2

Now, find the number of terms as follows:

[9,999,999,999 - 1,000,000,001] / 2 + 1 =4,500,000,000 terms that are divisible by 2. This makes sense, since half of the 9,000,000,000 numbers, which are 10-digit numbers,  are divisible by 2.

If P(x) is a cubic polynomial with leading coefficient 2 such that P(1) = 2, P(2) = 8, P(3) = 18, find P(4).

A triangle (ABC) is inscribed in a circle of radius 15.  The measures of the angles are in the ratio 2:3:4.  Find the lengths of the corresponding arcs.

Circumradius   R = 15

The ratio of the angles is   2  :  3  :  4                   40°    60°    80°

Arc measures are                                              80°       120°     160°

Arc lengths are                                 20.944           31.416           41.888    

A quarter-circle with radius 5 is drawn. A circle is drawn inside the sector, which is tangent to the sides of the sector, as shown. Find the radius of the inscribed circle.

I prefer to draw a graph first, for these sort of questions:

Find a monic quartic polynomial f(x) with rational coefficients whose roots include x=1-\sqrt 2 and x=2+\sqrt 5. Give your answer in expanded form.

\((x-(1-\sqrt{2}) )(x + 1-\sqrt{2} )(x-(2+\sqrt{5} ))(x+2+\sqrt{5})\)

\((x^2 -(1-\sqrt{2})^2)(x^2 - (2+\sqrt{5})^2)\)

\(x^4 -[(1-\sqrt{2})^2+(2+\sqrt{5})^2]x^2+(1-\sqrt{2})^2(2+\sqrt{5})^2\)

Can you finish it off from there?

Let a b c d and e be positive integers.

The sum of the four numbers on each of the five segments connecting "points" of the star is 28.

What is the value of the sum \(a+b+c+d+e\)?

I assume:

\(\begin{array}{|lrcll|} \hline (1): & e+4+b &=& 28 \\ (2): & a+5+c &=& 28 \\ (3): & b+6+d &=& 28 \\ (4): & c+7+e &=& 28 \\ (5): & d+8+a &=& 28 \\ \hline \text{sum}: & 2(a+b+c+d+e) + 4+5+6+7+8 &=& 5*28 \\ & 2(a+b+c+d+e) + 30 &=& 140 \\ & 2(a+b+c+d+e) &=& 110 \\ & \mathbf{a+b+c+d+e} &=& \mathbf{55 } \\ \hline \end{array}\)

Solve

\(|x - 1| - 2|x| = 3|x + 1|\).

\(\begin{array}{|rcll|} \hline \text{zero in }~ |x - 1|: \\ \hline x - 1 &=& 0 \\ \hline \mathbf{x} &=&\mathbf{ 1 } \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{zero in }~ |x|: \\ \hline \mathbf{x} &=&\mathbf{0} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{zero in }~ |x + 1|: \\ \hline \hline x + 1 &=& 0 \\ \hline \mathbf{x} &=&\mathbf{ -1 } \\ \hline \end{array}\)

We create a sign table:

\(\begin{array}{|l|c|c\|\|c|c|c|c|c|} \hline \text{Interval or position} : & \left(-\infty,-1\right) & -1 & \left(-1,0\right) & 0 & \left(0,1\right) & 1 & \left(1,\infty\right) \\ \hline \text{sign of } (x+1): & - & 0 & + & + & + & + & + \\ \hline \text{sign of } (x): & - & - & - & 0 & +& +& + \\ \hline \text{sign of } (x-1): &- & - & - & - & - & 0 & + \\ \hline \end{array} \)

For the 4 cases  \((-\infty, -1] ,\ (-1 , 0] ,\ (0, 1] ,\ (1, \infty)\)
we can therefore always resolve all amounts and solve the corresponding equation without amounts,
taking into account the respective case condition.
The negative terms from the amounts are highlighted in green:

\(\begin{array}{|rcll|} \hline (-\infty, -1] \\ \hline {\color{green}-(x - 1)} - 2({\color{green}-x}) &=& 3({\color{green}-(x + 1)}) \\ -x+1 +2x &=& -3(x + 1) \\ -x+1 +2x &=& -3x -3 \\ 4x &=& -4 \\ \mathbf{x} &=& \mathbf{-1}\ \checkmark \qquad -\infty < x \le -1 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline (-1 , 0] \\ \hline {\color{green}-(x - 1)} - 2({\color{green}-x}) &=& 3(x + 1) \\ -x+1 +2x &=& 3x+3 \\ 2x &=& -2 \\ x &=& -1 \quad \mathbf{\text{no}} \qquad -1 < x \le 0 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (0, 1] \\ \hline {\color{green}-(x - 1)} - 2(x) &=& 3(x + 1) \\ -x+1 -2x &=& 3x+3 \\ 6x &= & -2 \\ x &=& -\dfrac{1}{3} \quad \mathbf{\text{no}} \qquad 0 < x \le 1 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (1, \infty) \\ \hline (x - 1) - 2(x) &=& 3(x + 1) \\ x-1 -2x &=& 3x+3 \\ 4x &= & -4 \\ x &=& -1 \quad \mathbf{\text{no}} \qquad 1 < x < \infty \\ \hline \end{array}\)

\(\mathbf{x = -1}\)

Solve for x if 1/(x + 2) = 1/2 + 1/x.

One of their common tangents is touching as shown in the figure. 

Find the radius of the circle with center R.

\(\text{Let $PR=p+r$} \\ \text{Let $PB=p-r$} \\ \text{Let $QR=q+r$} \\ \text{Let $QC=q-r$} \\ \text{Let $PQ=p+q$} \\ \text{Let $PA=p-q$} \\ \text{Let $DF=AQ$} \\ \text{Let $DF=DE+EF$}\)

\(\begin{array}{|rcll|} \hline \mathbf{DE^2 +PB^2} &=& \mathbf{PR^2} \\ DE^2 + (p-r)^2 &=& (p+r)^2 \\ DE^2 &=& (p+r)^2 - (p-r)^2 \\ DE^2 &=& p^2+2pr+r^2-p^2+2pr-r^2 \\ DE^2 &=& 2*2pr \\ \mathbf{DE} &=& \mathbf{2\sqrt{pr} } \\ \\ \hline \mathbf{EF^2 +QC^2} &=& \mathbf{QR^2} \\ EF^2 + (q-r)^2 &=& (q+r)^2 \\ EF^2 &=& (q+r)^2 - (q-r)^2 \\ EF^2 &=& q^2+2qr+r^2-q^2+2qr-r^2 \\ EF^2 &=& 2*2qr \\ \mathbf{EF} &=& \mathbf{2\sqrt{qr} } \\ \\ \hline \mathbf{DF^2 + PA^2} &=& \mathbf{PQ^2} \\ DF^2 + (p-q)^2 &=& (p+q)^2 \\ DF^2 &=& (p+q)^2 - (p-q)^2 \\ DF^2 &=& p^2+2pq+q^2-p^2+2pq-q^2 \\ DF^2 &=& 2*2pq \\ \mathbf{DF} &=& \mathbf{2\sqrt{pq} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{DF} &=& \mathbf{DE+EF} \\ 2\sqrt{pq} &=& 2\sqrt{pr}+2\sqrt{qr} \quad | \quad : 2 \\ \sqrt{pq} &=& \sqrt{pr}+\sqrt{qr} \\ \sqrt{p}\sqrt{q} &=& \sqrt{p}\sqrt{r}+\sqrt{q}\sqrt{r} \quad | \quad : \sqrt{r} \\\\ \dfrac{\sqrt{p}\sqrt{q}}{\sqrt{r}} &=& \sqrt{p} +\sqrt{q} \quad | \quad : \sqrt{p}\sqrt{q} \\\\ \dfrac{1}{\sqrt{r}} &=& \dfrac{\sqrt{p}}{\sqrt{p}\sqrt{q}} +\dfrac{\sqrt{q}}{\sqrt{p}\sqrt{q}} \\\\ \mathbf{\dfrac{1}{\sqrt{r}}} &=& \mathbf{\dfrac{1}{\sqrt{q}} +\dfrac{1}{\sqrt{p}}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{1}{\sqrt{r}}} &=& \mathbf{\dfrac{1}{\sqrt{q}} +\dfrac{1}{\sqrt{p}}} \quad | \quad p=16,\ q=4 \\\\ \dfrac{1}{\sqrt{r}} &=& \dfrac{1}{\sqrt{4}} +\dfrac{1}{\sqrt{16}} \\\\ \dfrac{1}{\sqrt{r}} &=& \dfrac{1}{2} +\dfrac{1}{4} \\\\ \dfrac{1}{\sqrt{r}} &=& \dfrac{3}{4} \\\\ \sqrt{r} &=& \dfrac{4}{3} \quad | \quad \text{square both sides}\\\\ r &=& \dfrac{16}{9} \\\\ r &=& 1.\bar{7}\ \text{cm} \\ \hline \end{array} \)

The radius of the circle with center R is \(\mathbf{1.\bar{7}\ \text{cm}}\)

This is your answer:

\(\)\(5!-5\)

Thank you for posting.

Hey, question "equation of the line tangent to the circle at (-5, 12).".

I do not know what you are talking about.

Please repost this message with an image.

Not officially locked, but moderators can totally lock this.

Løcked and denied.

Edit1: no likes/hearts

Find constants  and  such that
\(\dfrac{x + 7}{x^2 - x - 2} = \dfrac{A}{x - 2} + \dfrac{B}{x + 1}\)

\(\begin{array}{|lrcll|} \hline & \dfrac{x + 7}{x^2 - x - 2} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad | \quad x^2 - x - 2 =(x - 2)(x + 1) \\\\ & \dfrac{x + 7}{(x - 2)(x + 1)} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad | \quad \times (x - 2)(x + 1) \\\\ & \mathbf{x + 7} &=& \mathbf{A(x + 1) + B(x - 2)} \\ \hline x=2: & 2 + 7 &=& A(2 + 1) + B(2 - 2) \\ & 9 &=& 3A + 0 \\ & 9 &=& 3A \\ & A &=& \dfrac{9}{3} \\ & \mathbf{A} &=& \mathbf{3} \\ \hline x=-1: & -1 + 7 &=& A(-1 + 1) + B(-1 - 2) \\ & 6 &=& 0 -3B \\ & 6 &=& -3B \\ & B &=& -\dfrac{6}{3} \\ & \mathbf{B} &=& \mathbf{-2} \\ \hline \end{array} \)

\(\dfrac{x + 7}{x^2 - x - 2} = \dfrac{3}{x - 2} - \dfrac{2}{x + 1}\)

Three men can paint a barn in 12 hours.  How long will it take two men to paint a barn?

Hello Guest!

\(\color{blue}W=N\cdot t\cdot n\\ N=\frac{W}{t\cdot n}=\frac{W}{12h\cdot 3}\)

\(N=\frac{W}{36h}\)

\(W=N\cdot t\cdot n\\ W=\frac{W}{36h}\cdot t\cdot 2\\ t=\frac{W\cdot 36h}{2W}\)

\(t=18h\)

Two men can paint a barn in 18 hours.

  !

X², your presentations are always graceful works of detailed art; made understandable with your grace notes.  

You’re Amazing! 

 ^{_{For sure, none your presentations are ever a waste of time! There are many who appreciate your work and you.}}

GA

The answer is a = -2.

Solution by computer program:

count = 0;

for (n = 1000000000; n <= 9999999999; n = n + 1) {

  if (sumofdigits(n)  % 2 == 0) {count = count + 1;}

}

write(output,count);

output = 2,000,000 numbers

The radius is 5*sqrt(2)/2.

If you let x go infinity, then you get 0 = A + B.

If you let x = 0, then you get -7/2 = A/(-2) + B.

The solution to this system is then A = 7/3, B = -7/3, so (A,B) = (7/3,-7/3).