CrimsonLots503

Had there not been any limitations; we could've done the following:

First, we can start with some basic rearrangements.

\(2x^2 - 10x + 13 = -6x^2 - 18x - 15\),

\(8x^2 + 8x + 28 = 0\), 

\(x^2 + x + \frac{7}{2} = 0 \)

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We turn this into a monic quadratic to simplify our working out. Since the roots of the quadratic are \(a+bi, a-bi\)

, we can use Vieta's formula to see that:

\((a+bi)+(a-bi) = 2a = -1 \rightarrow a = -\frac{1}{2}\)
\((a+bi)\cdot(a-bi) = a^2 + b^2 = \frac{7}{2}\)

We can substitute our value for a and get \(a^2 + b^2 = \frac{7}{2} \rightarrow (-\frac{1}{2})^2 + b^2 = \frac{7}{2} \rightarrow b^2 = \frac{13}{4} \therefore b = \frac{\pm\sqrt{13}}{2}.\)

Which means \(a\cdot b = -\frac{1}{2}\cdot\frac{\pm\sqrt{13}}{2} = \frac{\pm\sqrt{13}}{4}.\)

The value for a can't be positive for the solutions for that equation. Since the roots are literally \(\frac{-1\pm\sqrt{13}i}{2}\). b > 0 but not a.

We can do some rearrangements to do the following:

\(\frac{a + 2}{a + 1} = \frac{b}{3}\), 

\(\frac{3a + 6}{a + 1} = b\), 

\(\frac{3a + 3}{a + 1} + \frac{3}{a + 1} = b\)
\(3 + \frac{3}{a + 1} = b\).

From this, we have to find values of a for which \(\frac{3}{a + 1}\) is an integer. Since there's no limitation on positive or negative integers, we find that if
\(a = 2 \rightarrow b = 4\)
\(a = -4 \rightarrow b = 2\)

\(100^{100} = (10^2)^{100} = 10^{200}.\)
Now we can check the remainder of the powers of 10 being divided by 18.

\(10^2 \equiv 10 mod(18) \because 10^2-(18*5) = 10\)

\(10^3 \equiv 10 mod(18) \because 10^3-(18*55) = 10\)

\(10^4 \equiv 10 mod(18) \because 10^4-(18*555) = 10\)

\(10^5 \equiv 10 mod(18) \because 10^5-(18*5555) = 10\)

\(10^6 \equiv 10 mod(18) \because 10^6-(18*55555) = 10\)

...

Therefore we can conclude that the remainder of \(100^{100}\) divided by 18 is 10

A rhombus with an angle of \(90^\circ\) is the same as a square. Since the sides of a rhombus are all of equal length and the opposing angles are equal. If one of the angles are \(90^\circ \)then all of the angles are \(90^\circ \). Which means the area is \(12^2 = 144\)

I just want to add on that we can represent the answer \(10\sin(75) =  10\tan(15) = 10\tan(\frac{30}{2}).\)
Now we can use the trigonometric half-angle identity \(\tan(\frac{\alpha}{2}) = \frac{\sin(\alpha)}{\cos(\alpha)+1}\).
We can input our normal values: \(10\tan(\frac{30}{2}) = 10(\tan(\frac{30}{2})) = 10(\frac{\sin(30)}{\cos(30)+1}) = 10(\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}+1) = 10(\frac{1}{\sqrt{3}+2}) = 10(2-\sqrt{3})= 20 - 10\sqrt{3}.\)
Therefore \(BE = \sqrt{(20 - 10\sqrt{3})^2+10^2} = \sqrt{400 + 300 - 400\sqrt{3} + 100} = \sqrt{800 - 400\sqrt{3}} = \sqrt{400(2-\sqrt{3})} = 20\sqrt{2-\sqrt{3}}\)

1. Our most obvious step is to draw the points \(O, P, Q\) and label their lines' respective gradients: \(OP_m = 4, OQ_m = 5\). We are trying to find out \(PQ_m\). Where m stands for the gradient. \([y = mx + c]\)
2. Let's label point Q(a, 5a) and P(b, 4b). Since OP = OQ, that means we can equate their sum of squares together.

3. \(b^2 + (4b)^2 = a^2 + (5a)^2\)

  i) \(b^2 + 16b^2 = a^2 + 25a^2\)

  ii) \(17b^2 = 26a^2\)

  iii) \(b\sqrt{17} = a\sqrt{26}\)

  iv) Now we can express b in terms of a and produce \(b = a\sqrt{\frac{26}{17}}\)

4. To find \(PQ_m\), we need to use our gradient formula which is \(\frac{y_2-y_1}{x_2-x_1} = \frac{4b-5a}{b-a}\)

5. for simplification we see that \(\frac{4b-5a}{b-a} = \frac{4b-4a-a}{b-a} = \frac{4(b-a)-a}{b-a} = 4 - \frac{a}{b-a}\)

6. Now we can express a in terms of b which turns \(\frac{a}{b-a} = \frac{a}{a\sqrt{\frac{26}{17}}-a} \)

7. \(\frac{a}{a(\sqrt{\frac{26}{17}}-1)} = \frac{1}{\sqrt{\frac{26}{17}}-1)},\) which we rationalize by multiplying both the numerator and denominator by\( \sqrt{\frac{26}{17}}+1\) which gives us \(\frac{\sqrt{442}+17}{9}\)

8. Therefore our gradient is \(4 - \frac{\sqrt{442}+17}{9} = \frac{19-\sqrt{442}}{9}\)
 

I'm pretty sure this slope is negative so I'm confident this may be the right answer. However, please do check for any calculation errors I've made :)

Let's assign the values of the dimensions of the prims as \(p, q, r.\) 
For the surface area: \(2(pq + pr + qr) = 15\)
For the sides: \(p + q + r = \frac{17}{4}\)
To find the diagonal we have to know: \sqrt{p^2 + q^2 + r^2}

With some basic algebraic manipulation, we know that \((p+q+r)^2 = p^2 + q^2 + r^2 + 2(pq + pr + qr)\), therefore:

\(p^2 + q^2 + r^2 = (p+q+r)^2 - 2(pq + pr + qr)\). We can substitute the values we know and write, 
\(p^2 + q^2 + r^2 = (\frac{17}{4})^2 - (15), p^2 + q^2 + r^2 = \frac{289}{16} - \frac{240}{16},  p^2 + q^2 + r^2 = \frac{49}{16}, \)

therefore \(\sqrt{p^2 + q^2 + r^2} = \sqrt{\frac{49}{16}} = \frac{7}{4}\). 

Now we found out that the diagonal of the prism is \(\frac{7}{4}\) :)