All Answers 358108 Answers

Simplify the function first.

\(g(x) = \dfrac2{2 + 4x^2 + 5 + 9x^2} = \dfrac2{13x^2 + 7}\)

Note that 13x^2 + 7 is always positive. So g(x) is always positive.

Also note that the minimum of 13x^2 + 7 is attained at x = 0, with a minimum value of 7.

Therefore, \(\displaystyle\max_{x\in\mathbb R} g(x) = \dfrac2{\displaystyle\min_{x\in\mathbb R}(13x^2 + 7)} = \dfrac27\).

Also, note that \(\displaystyle\lim_{x\to +\infty} g(x) = \displaystyle\lim_{x\to -\infty} g(x) = 0\).

Therefore, the range is \(\left(0, \dfrac27\right]\).

Note that \(13 + 2(17) = 47\).

Multiply both sides by 29 gives 

\(13 \times 29 + 2 \times 17 \times 29 = 47 \times 29 \equiv 0 \pmod{47}\\ 1 + 17 \times (2 \times 29) \equiv 0 \pmod {47}\\ 17 \times (-2\times 29) \equiv 1\pmod{47}\\ 17 \times (-58) \equiv 1 \pmod{47}\\ 17 \times 36 \equiv 1 \pmod{47} \)

Therefore, by definition of modular inverse, \(17^{-1} \equiv 36 \pmod{47}\).

Consider the prime factorization of the answer. The factorization cannot contain primes smaller than 20.

That means all prime factors must be greater than or equal to 20.

The smallest prime greater than or equal to 20 is 23.

Since it cannot be just 23 (it would be a prime, not a composite number), we arrive at the conclusion that \(23^2 = 529\) is the smallest composite number that has no prime factors less than 20.

So, 

\(\begin{cases}x + y = 3\\xy = -18\end{cases}\)

You can check by calculating the sum and product that x = 6, y = -3 is a solution.

Alternatively, if you know how to solve a quadratic equation, by Vieta's formula, x and y are the roots of \(t^2 - 3t - 18 = 0\).

So \(f(x) \neq 0\) and \(x \geq 0\) since \(x^{1/2} = \sqrt x\).

\(f(x) \neq 0\\ 2 - x^{1/2} \neq 0\\ x \neq 4\)

Combine \(x \geq 0\) and \(x \neq 4\) to get \(x \in [0, 4) \cup (4, +\infty)\).

Therefore, the domain of g(x)/f(x) is \([0, 4) \cup (4, +\infty)\), or written in inequality form, \(\{x \in \mathbb R : 0 \leq x < 4 \text{ or }x > 4\}\).

Note that when x = -2 or x = 2, the denominator will become 0.

Things inside any square root must be nonnegative and the denominator is nonzero.

\(x^2 - 4 \geq 0\text{ and } \sqrt{x^2 - 4} \neq 0 \\ x^2 - 4 \geq 0 \text{ and }x^2 - 4 \neq 0\\ x^2 - 4 > 0\)

Continue to solve the inequality, and you will get the domain of the function. 

\(\operatorname{Domain}(f) = \{x \in \mathbb R:-2 < x < 2\}\)

-3(x - 3) > 5 - 5x

3(x - 3) + 5 - 5x < 0

-2x - 4 < 0

-2x < 4

Now please finish it. Be careful that when you divide by a negative number on both sides of an inequality, you need to invert the inequality sign.

Suppose that \(y = r + x\), where \(0 \leq x < 1\) and r is an integer.

Then \(\lfloor y \rfloor = r\).

\(r(r +x) = 42\\ x = \dfrac{42}r - r\)

Then x has to satisfy the inequality \(0 \leq x < 1\).

We find the range of values of r such that \(0 \leq \dfrac{42}r - r < 1\). 

Since y > 0, r >= 0. We multiply each term by r to get \(0 \leq 42 - r^2 < r\).

Solving \(0 \leq 42 - r^2\) gives \(-\sqrt{42} \leq r \leq \sqrt{42}\).

Solving \(42 - r^2 < r\) gives \(r > 6\text{ or }r < -7\).

Combining the solutions of the two inequalities gives \(6 < r \leq \sqrt{42}\). But this inequality has no integer solutions.

Therefore, there are no such value of y such that y * floor(y) = 42.

Suppose x <= -7. When f(x) = 0, -x - 3 = 0. This means x = -3, which is not <= -7. This root is rejected.

Suppose x > -7. When f(x) = 0, 2x + 1 = 0. This means \(x = -\dfrac{\boxed{\phantom{\text{aaaaa}}}}{\boxed{\phantom{\text{aaaaa}}}}\).

Therefore, the sum of all values of x such that f(x) = 0 is \(-\dfrac{\boxed{\phantom{\text{aaaaa}}}}{\boxed{\phantom{\text{aaaaa}}}}\), as it is the only solution to f(x) = 0.

Please fill in the blanks.

Evaluate each of the terms separately. Note the formulas \(\sqrt{\dfrac ab} = \dfrac{\sqrt a}{\sqrt b}\) and \(\left(\dfrac ab\right)^2 = \dfrac{a^2}{b^2}\). Please try to take it from here.

Oooh!!! I get it now!!!

Thank you very much!!!!!! 

Your 5C1 gives you   5 choices    DLRIE   

If you chose  L  then your 4C1 will give you the 4 coices   DRIE   say you then chose D

you now have     COL    COD    RIE

alternatively you could have first chosen  D   and then chosen  L   

then you would have  COD   COL   RIE

But these are the same, you have counted this outcome twice.

Yes it is the same but how would this occur (how would I have counted it twice If i used 5C1 then 4C1) 

Maybe I do not understand your question but

COD   COL   RIE      is the same as   COL   COD    RIE     You have counted it twice.

I think 65 is correct.

Responded; thanks!

sin( theta) is positive becasue it is in the second quadrant

Find the opposite side using Pythagoras's Theorem.  Then read of sin (theata)

For 1):
But how would COD COL *** and COL COD *** occur?

To clarify: in the first space, we have 5 letters and we want to select 1 letter so 5C1 (This will give one of the following letters: R D I L E) Suppose it gave D. Then, in the second space, we have 4 letters now (R I L E) and we select 1 (for example, L) so 4C1, and the last 3 letters ( R I E ) in the last group.

So, how would C O L C O D  *** appear ? ( I think It would appear if I said 5C1 and 5C1 again (5C1*5C1) but I removed a letter (5--->4))

By the way, 65 was also an answer I got randomly by other methods (but can't see the logic behind these, also the answer 105 and 210,420 etc.. all have some mistakes such as not accounting for the identical "Os" , and for 65 the above ^ reason is what I do not understand yet, I mean if it is possible that COD COL *** and COL COD *** would occur by case (1) then definitely, 65 is the correct answer! but I do not see how would these both occur at the same time by the same case.

anyway, this question is really silly :( ... 

Thank you :)!!

I would like you to become a member.

We need people who are seriously trying to understand :)

If you get a good reputation your questions will get priority.

I have added my 2c worth on the original thread :)

mmm interesting use of LaTex.

I will try to talk of your solution.    You have made a very good attempt    

1)   CO _    CO_    _ _ _        You got 20 here.  I think you need to divide that by two becasuse

                                            for example COD COL ***      is the same as   COL COD ***

                                             so I think this should be 10

2)   COO  C_ _    _ _ _       You got 10, I agree

3)    CO_   C_ _   O_ _         We agree on 30

4)   C_ _   C_ _   OO_          You got 30 I would halve that for the same  reason as in 1.         So 15 I think.

I think that covers all the cases so I get     10+10+30+15 = 65  ways.

I am not going to guarentee that it is correct though   

Ok, take your time! 

ok, I have had a bit of a look, it is not an easy one.

I will look some  more.    

thats my attempt..

There is already an answer there ??