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I'd look at the 4 cases individually   (pos can include zero I guess)

x and y positive

x and y both negative

x pos and y neg

y pos and x neg.

This is amazing. I just posted an answer to this clock problem on a more recent posting of the problem. I had not seen Phil's solution, but I could have believably copied his verbatim. The same thing happened on a regular decagon problem, even though our solutions were not quite so similar. This is fun;  post problems as many times as your heart desires, even though I am getting dizzy looking through the solutions. Just kidding.  Please have mercy. I really think only one posting should be allowed at one time, so no time is wasted and the better solutions can be compared.By the way, I tried to sign up today, but the activation email never arrived. Still waiting patiently.

ok Sorry.

This time I will close it again.

You are asking multiple questions and it seems you are trying to disguise this fact by posting as 2 different usernames AND a guest.

One person one name. 

Go back and ask again under your earlier name.

Let the tangent to the circle be TPR then, 

∠OTP = ∠QPR = 28°         ...[PQ∥TO and corres. angles] 

OP ⊥ TR       ...[TR is a tangent and OP is the radius] 

⇒ ∠OPR = 90°

 ∠OPQ = ∠OPR -  ∠QPR = 90 - 28 = 62

 ∠OPQ = 62° 

∵ OP = OQ           ...[radii of circle] 

⇒ ∠OPQ = ∠OQP = 62°

In △OPQ

 ∠OPQ +  ∠OQP + ∠POQ = 180 

                               ∠POQ = 180 - 124 

                               ∠POQ = 56°

At 10 o'clock the hour and minute hands are 60 degrees apart.

In one minute the minute hand advances \(\frac{1}{60}(360)=6\) degrees.

The hour hand, on the other hand (lots of puns that should be intended), advances \(\frac{1}{60}(30)=0.5\) degrees.

So the angle between the two hands increases by \(6-0.5=5.5 \)degrees per minute.

We are looking for the time when the angle is 90 degrees, that is when the angle has increased by 30 degrees.

It takes \(\frac{30}{5.5} \)or approximately 5.45 minutes for that to happen. So the two hands are at precisely 5 minutes and 27 seconds after 10  perpendicular to each other.

I don't have a clue. Sorry.

How about doing some of your own homework?

I'd written quite a lengthy answer to this but then the page accidently reloaded and damnit there goes efforts :/ but thankfully I'm gifted with extraordinary patience so I decided to give it another go. 

So the point on the line \(-6x-5y+1=0\) which is closest to the point (3,1) is the perpendicular/normal on that line from that particular point, as we know that the normal is the shortest length from any point right? 

Equation of line is 

 \(-6x-5y+1=0\)          ...(1)

Slope \(={-5\over 6}\)

Slope of normal \(={-1\over {-5\over 6}} = {6\over 5}\)

Now we find the equation of that normal from the point (3,1)

\(y-1={6\over 5}(x-3)\)

\(5y-5=6x-18\)   ⇒\(6x - 5y-13=0 \)      ...(2) 

Now that point is where the two lines intersect so we equate eq(1) and (2) 

\(-6x-5y+1=6x-5y-13\)

⇒ \(x= {7\over 6}\)   and   \(y={-6\over 5}\)

∴The point is \(({7\over 6},{-6\over 5})\). 

To wrap things up, I've also enclosed a graph here

I hope its clear. 

P.S. Turns out, I've explained this answer lot better than the previous one lol. If anything I've gained from writing this answer twice, its that I've gained a lot more patience that I already have 

There are 2 questions there.  Ask one on a new post and work the other out from any answer that you get.

I am closing this post.

Solve x^2=5   are either of the answers in the rage x<=-2?

Solve |2x+1|=5 is the answer/s in the required range?

Solve 3x+8=5  is the answer/s in the required range?

1) Find or state all the initial conditions.

What is the height?

the vertical speed and the horizontal speed?  

and the acceleration in vertical and horizontal directions?

To find the initial vertical and horizontal speeds.

Draw a right-angled triangle.  The right angle and 30 degrees will be at the base

The hypotenuse will be 70feet/sec

Use trig to find the initial vertical and horizontal speeds.

When you have done that get back with the worked answers (or explain your confusion)

I'd help if I could but I do not have a clue.....

Let f(x) = ax² + bx + c

⇒f(0) = c = 3

f(1) = a + b + c = -5

           ⇒ a + b = -8          

f(2) = 4a + 2b + c = 7 

           ⇒ 4a + 2b = 4   ⇒ 2a + b = 2

⇒ 2a + b - a - b = 2 + 8 

       a = 10

       b = -18

f(x) = 10x² - 18x + 3

In 100 gallons, alcohol = 50%

25 gallons is replaced with 20% alcohol

So, resulting sol \(=(75×{50\over 100})+(25×{20\over 100})\)

                          \(=37.5+5\)

                          \(=42.5\)% alcohol

The process is done again,

Resultant sol \(=(75×{42.5\over 100})+(25×{20\over 100})\)

                      \(=31.875+5\)

                      \(=36.875\)% alcohol

After the process is repeated again,

Resultant sol \(=(75×{36.875\over 100})+(25×{20\over 100})\)

                     \(=27.65625+5\)

                      \(=32.65625\) \(=33\)% alcohol

∴The final solution has 33% alcohol 

Shortly after 10 'o clock, the hands of a clock made a 90-degree angle. What's the time?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

60 - x/12 + x = 90          x = 32⁸/₁₁ degrees       ( x = 360 / 11 )

For brevity, all congruences below are taken modulo $n$.  $(ab)^{-1}\equiv 2$ means $2(ab)\equiv1$.  So we have $(2a)b\equiv a(2b)\equiv1$, meaning both $a$ and $b$ are invertible with $a^{-1}= 2b\bmod n$ and $b^{-1}=2a\bmod n$.  Thus,
\begin{eqnarray*}
    (a+b)^{-1}(a^{-1}+b^{-1}) &\equiv& (a+b)^{-1}(2b+2a) \\
    &\equiv& ((a+b)^{-1}(a+b))2 \\
    &\equiv& 1\cdot 2 \\
    &\equiv& 2.
\end{eqnarray*}

f(f(x)) = x for all x is a condition that says the function f is its own inverse.

We know that a function that is its  own inverse must have a graph that is symmetric with respect to the line y =x.

The red graph is a plot of the function f(x) = 8 -x, x<=4. If we reflect the red graph in the line y =x (the faint dashed line) in order to make it symmetric with respect to it, we get the blue graph.

The red and the blue graph have the same equation, but different domains; the blue graph is defined for values of x to the right of 4, whereas the red graph has values of x to the left of 4 as domain.

Since the two pieces of the function have the same equation we must have

                                                                                 f(x) = ax +b = 8 - x = -x +8,

that is, a = -1 and b =8, and thus a + b = -1 + 8 = 7.

1st Case: For cast iron                                                  2nd Case: For Al 

F = 6.5 × 10³ N                                                               F = 6.5 × 10³ N 

l = 1.3 m                                                                          l = 2.6 m

r = 0.2 × 10^-2 m  ⇒ A =  0.04π × 10^-4 m²                       A =  0.04π × 10^-4 m²

Y = 100 × 10¹⁰ N/m²                                                          Y = 7 × 10¹⁰ N/m²

Now, formula of Young's modulus is 

    \(Y={1\over 2}{Fl\over A△l}\)                                                                 \(△{l}_{2} = {1\over 2}{FL\over AY}\)

⇒\(△{l}_{1} = {1\over 2}{FL\over AY}\)                                                                        \(={1\over 2}×{6.5×10^3×2.6\over 0.04π×10^{-4}×7×10^{10}} \)

          \(={1\over 2}×{6.5×10^3×1.3\over 0.04π×10^{-4}×100×10^{10}} \)                                        

In order to get the elongation of the whole rod, calculate △l₁ and △l₂ separately and add them. 

∴ Elongation of rod = △l₁ + △l₂

I hope its clear :) 

Let \(n\) be a positive integer greater than or equal to \(3\).

Let \(a,b\) be integers such that \(ab\) is invertible modulo \(n\) and \((ab)^{-1} \equiv 2 \pmod{n}\).

Given \(a+b\) is invertible, what is the remainder when \((a+b)^{-1}(a^{-1}+b^{-1})\) is divided by \(n\)?

-2(6 + 2w) ≤ -16             and            -3w ≥ 18 

⇒ -12 - 4w ≤ -16                                  w ≤ -6

⇒ -4w ≤ -4 

        w ≥ 1

Now value interval of w = (-∞, -6] ∪ [1,∞)

∴ w doesn't exist in the interval = [-6,1]

Multiply \({(a+b)}^{-1}({a}^{-1}+{b}^{-1})\) by ab

\({(a+b)}^{-1}({a}^{-1}+{b}^{-1})\)(ab) =\({(a+b)}^{-1}({a}^{-1}(ab)+{b}^{-1}(ab))\)

                                             \(={(a+b)}^{-1}(b+a)\)

                                            = 1 (mod n)

This shows that \({(a+b)}^{-1}({a}^{-1}+{b}^{-1})\) = \({(ab)}^{-1}\)= 2 (mod n)

Suppose p is the greatest common divisor of n and 3n+4. We will show that p would have to be a factor of 4 with possible values 1, 2, and 4.

We have                                                  n = kp,                            (1)

and                                                                 3n+4= mp,                      (2)

for some. integers k, and m.

We also know that k and m have no common factors other than 1 (they are relatively prime) because of p being the greatest of the common factors.

Substitute (1) in (2) to get

                                                                       3(kp)+4 = mp                 (3)

Rearrange the terms in (3) to isolate the term '4' and factor out p:

                                                                       4 = p(m - 3k)                  (4)

Equation (4) shows that we have factored 4 as p times some nonzero integer (m cannot equal to 3times k), which shows that

p is a factor of 4. But positive integer factors of 4 are only 1, 2, and 4. Sum of these numbers is 7.

In theory, the approach is good, but as I said above, I don't think this will work.  

Okay, start with your        x² + (101 – x)²  =  327  

                                        x² + (10,201 – 202x + x²)  =  327   

                                        2x² – 202x + 10,201  =  327  

                                        2x² – 202x + 9874  =  0  

                                          x² – 101x + 4937  =  0  

                                          4937 is prime so we can't factor this.  

                                          The alternative is to use the quadratic formula.  

                                          I'll spare you the gory details, but you end up with a negative number under the radical.  

                                          I don't know where to go from here, mnkmnk.  Have I made a mistake?     

The roots of the cubic are 5, -12, and 9, and you would find (x, y, z), and from that x+y, by assigning a permutation of the roots to (3x, 5y, -6z). Three numbers can be permuted in 6 different ways, so there are 6 different assignments:

(3x, 5y, - 6z)=(9, 5, -12)

                   =(9, -12, 5)

                   = (5, 9, -12)

                   =(5, -12, 9)

                   = (-12, 5, 9)

                   =(-12, 9, 5)

You would find 6 values for x +y this way, and I think they turn out to be distinct.