Tiggsy

The roots of the equation are x^3 = 1 are 1, -(1/2) - I root(3)/2  and -(1/2) + I root(3)/2.

x =1 is excluded so x is one of the other two.

It doesn't matter which, since one is the square of the other.

What happens after that ....... .

Here's a non calculus approach.

Begin by noticing that

\(7^{2}+24^{2}=49+576=625=25^{2},\)

so 7, 24, 25 can be used to as the sides of a right-angled triangle.

Let the angle opposite the 7 be \(\phi,\)

then

\(\sin\phi=7/25 \text{ and } \cos\phi=24/25.\)

\(\displaystyle 7\cos\theta+24\sin\theta+7\\ = 25(\frac{7}{25}\cos\theta+\frac{24}{25}\sin\theta)+7\\ =25(\sin\phi\cos\theta+\cos\phi\sin\theta)+7 \\ =25\sin(\theta+\phi)+7.\)

Maximum value will be 25 + 7 = 32 occuring when

 \(\sin(\theta+\phi)=1, \\ \theta=\pi/2-\tan^{-1}(7/24),\\ 0\leq\theta\leq\pi/2.\)

Use the trig identities

\(\displaystyle \cos(A)+\cos(B)=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \\ \text{and} \\ \displaystyle \sin(A)+\sin(B)=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right).\)

The common 

\(\displaystyle \cos\left(\frac{A-B}{2}\right)\)

term will be equal to \(\cos(\pi/3)=1/2,\)

cancelling out the 2's.

\(\displaystyle (a-b)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3},\)

so subtracting the second equation from the first,

\(\displaystyle (a-b)^{3}=679 -615=64,\)

so

\(\displaystyle a-b=4.\)

\(\displaystyle \frac{2^{3}}{.2}+0!\)

(1 - 20x) ^(-1/2) = 1 + 10x + 150x^2 + 2500x^3 +...

The wording of this question strongly suggests that the use of Descartes' Rule of Signs is what's intended.

Suppose that we have a polynomial p(x) with real coefficients.

The rule says that the number of positive real roots is equal to the number of sign changes (as we move left to right along the polynomial) or is less than that number by an even integer.

Replacing x by -x to get the polynomial p(-x), a similar statement can be made about the number of negative real roots.

For example, the polynomial

\(\displaystyle p(x) = 3x^{5}-2x^{4}+7x^{2}-9 \\ p(-x)=-3x^{5}-2x^{4}+7x^{2}-9,\)

will have 3 or 1 positive and 2 or 0 negative real roots.

Thankyou for the feedback GA, ..... and I shall endeavour to find out what hypergeometric distribution counting is.

Tiggsy

Here are my attempts, comments/refutations welcome.

First, a counting type approach.

The pack consists of 28 cards, (4 card colours and 7 numbers for each colour).

The number of 5 card combinations that can be drawn will be C(28,5).

Needed are 3 cards with the same number, so 7*C(4,3) = 28 possibilities.

Assuming that the possibility of having 4 cards with the same number is excluded, the remaining 2 cards have to come from 24 remaining cards, (4 cards each of the other 6 numbers), C(24,2) possibilities.

That generates a probability of     \(\displaystyle \frac{28*C(24,2)}{C(28,5)}=\frac{46}{585} \approx 7.86 \%.\)

Second, a probability type approach.

Choose a number and suppose that we draw, in succession, three cards with that number, followed by any two cards with a different number.  

The probability of that sequence will be 

\(\displaystyle \frac{4}{28}\times \frac{3}{27} \times \frac{2}{26}\times \frac{24}{25}\times \frac{23}{24} =\frac{23}{20475}\)

Now there are 7 different numbers to choose from and the sequence can occur in 10 different ways, so the final probability will be

\(\displaystyle 7 \times 10 \times \frac {23}{20475}= \frac{46}{585} \approx 7.86 \%, \text{as earlier}.\)

Good luck throwing a 1 and a 12.