All Answers 358923 Answers

Thanks Ginger   

I do not really understand your question, I think it is incomplete but perhaps you only want this part of it.

Given f(n)=1/(n+1)+1/(n+2)+...+1/(3n+1). When you want to prove the question using the mathematical induction method. For step 2, if n=k, you want to validate when n=k+1, then the relationship between f(k) and f(k+1) will be?

\(f(1)=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\\ f(n)=\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}+ ....+\frac{1}{3n}+\frac{1}{3n+1}\\ f(k)=\frac{1}{k+1}+\frac{1}{k+2}+....+\frac{1}{2k}+ ....+\frac{1}{3k}+\frac{1}{3k+1}\\~\\ f(k+1)=\frac{1}{k+2}+.......\frac{1}{2k+2}+......\frac{1}{3k+3}+\frac{1}{3k+4}\\ f(k+1)=\frac{1}{k+2}+.....+\frac{1}{3k+1}+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}\\ f(k+1)=\left[\frac{1}{k+2}+.....+\frac{1}{3k+1}\right]+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}\\ f(k+1)=\left[f(k)-\frac{1}{k+1}\right]+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}\\ f(k+1)=f(k)+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}-\frac{1}{k+1}\\ \)

check:

\(f(1)=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\\ f(2)=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\\ f(2)=f(1)+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{2}\)

Well that looks promising.  :)

Laughter is usually among the best of medicines; although, I suppose it could top-off someone on the edge.  Anyway, you’ve had a few maga-doses and it’s not topped you yet, so keep borrowing from the borrowed-time zone –the interest rates are cheap, and though the days are getting shorter, you’ll have more of them, and you’ll make it up in volume.  

What’s the “other way” that might be better? 

GA

^{Just picture all the 2's and 4's with extra dots turning them all into 6's, say.}

^{To all intents and purposes they're the same, same as the 1's, 3's and 5's.}

Twos, fours, and sixes, Oh MY! This sounds like analysis using magical incantations and charms. “Snake eyes and boil, slick lies and toil, these ducks are now pigs.”  The ducks aren’t really pigs, but some members of the viewing audience may think they are.  This is entertaining, but not as much as using telekinetic powers to roll snake eyes or not snake eyes.

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DC Thomas, my cat, demonstrated his telekinetic powers for rolling snake eyes. This was very impressive to watch. He rolled snake eyes on about 60% of the attempts. The dice seemed to roll with normal expectation when he wasn’t using his telekinetic powers. This went on for a while, until Mr. Peabody, my dog, caught him switching out the loaded dice. The fight was on...

DC knows he’ll get caught when he uses loaded dice playing craps, or embezzles Monopoly money when he’s the banker. He does this just to pissoff Mr. Peabody –he likes to fight. Mr. Peabody always obliges. Long ago, I started securing the fragile artwork, and put the veterinarian on speed-dial, whenever we play games.

GA

Hi Melody,

The (4!) is in there for the same reason it’s in your solution. Four of the dice are indistinguishable from each other, so their permutations are factored out of the (10!).

To clarify, though the four dice appear to be distinguishable from each other because 2, 4, and 6 are distinguishable, they are treated as an indistinguishable group, that is, they are only distinguishable from the 1, 3, and 5. So this question has four groups: three groups of 2 indistinguishable element, each consisting of 1s, 3s, and 5s, and one group four that is an indistinguishable collective of NOT 1s, 3s, or 5s.

GA

It is good to see you back Max 

Let r(p) be the revenue.

r(p) = p(128-4p) = -4p^2 + 128p

Completing the square, r(p) = -4(p^2-32p) = -4(p-16)^2 + 1024.

Therefore, if the price is 16 dollars, the revenue will be maximized, which will be 1024 dollars.

Assuming you mean f(0), f(-2),f(3) instead of F being the antiderivative of f.

f(0) = 3(0) - 4 = -4

f(-2) = 3(-2) - 4 = -10

f(3) = 3(3) - 4 = 5

Now assume you mean the antiderivative.

F(0) = 3(0)^2 / 2 - 4(0) = 0

F(-2) = 3(-2)^2 / 2 - 4(-2) = 14

F(3) = 3(3)^2 / 2 - 4(3) = 3/2

Range of f(6x) is [-11,3]. So range of g(x)=f(6x)+1 is [-10,4]

\(r(x) = \dfrac{1}{(1-x)^2}\)

When x -> 1, r-> infinity. So the maximum value of r(x) is +infinity.

Now we just have to find the minimum value of r(x).

Horizontal asymptote of r(x) is y = 0, which means the r(x) is approaching 0 when x approachs +inf or -inf.

Therefore range of r(x) is (0, inf).

"In right triangle ABC with angle B = 90˚" implies that 0˚ < A < 90˚.

2 sin A = 3 cos A

2 tan A = 3

tan A = 3/2

A = arctan(3/2) = arccot(2/3)

sin A = sin(arccot(2/3)) = 1/sqrt(csc^2(arccot(2/3)) = 1/sqrt(1+cot^2(arccot(2/3))) = 1/sqrt(1+4/9) = \(\dfrac{3\sqrt{13}}{13}\)

We need to solve \(2(2^x-1)x^2+(2^{x^2}-2)x-2^{x+1}-2=0\).

Let \(f(x) = 2(2^x-1)x^2+(2^{x^2}-2)x-2^{x+1}-2\).

If you graph f(x) using any method, you find that it only touches x-axis at one point.

Graph link: https://www.desmos.com/calculator/wxie170yvi

We will use Newton-Rhapson to find solutions.

\(f'(x) = 4x(2^x-1)+2x^22^x\ln2+2^{x^2}-2+2x^2(2^{x^2}\ln2)-2^{x+1}\ln2\)

As this is too complicated, I solved it with a program.

Here is the source code in C++:
#include
using namespace std;
double f(double x){
    return 2*(pow(2,x)-1)*x*x+(pow(2,x*x)-2)*x-pow(2,x+1)-2;
}
double f_prime(double x){
    return 4*x*(pow(2,x)-1)+2*x*x*pow(2,x)*log(2)+pow(2,x*x)-2+2*x*x*(pow(2,x*x))*log(2)-pow(2,x+1)*log(2);
}
int main(){
    double x = 1.1;
    for (int i=0;i<1000;i++){ //iterate 1000 times
        x -= f(x)/f_prime(x);
    }
    printf("Solution: %100g", x);
}

Output: Solution: 1.32006

First we will need to find g(x).

\(g(3x-8) = 2x^2+5x-3\\ \quad g(x) \\= g\left(3\left(\dfrac{x+8}{3}\right)-8\right) \\= 2\left(\dfrac{x+8}{3}\right)^2+5\left(\dfrac{x+8}{3}\right)-3\\ =\dfrac{2}{9}\left(x^2+16x+64\right)+\dfrac{5}{3}(x+8)-3\\ =\dfrac{2}{9}x^2 +\dfrac{47}{9}x+\dfrac{221}{9}\)

Therefore g(-5) = \(\dfrac{2}{9}(-5)^2+\dfrac{47}{9}(-5)+\dfrac{221}{9}=\dfrac{50-235+221}{9} = \dfrac{36}{9} = 4\)

Ans = 4.

From \(g(3) = 9\), we know that \(f(9) = 3\).

From \(g(0) = 3\), we know that \(f(3) = 0\).

f(f(9)) = f(3) = 0.

Other informations are useless ones.

\(a\spadesuit 3 = 9\\\dfrac{a^2}{3} = 9\\a^2 = 27\\a = \pm\sqrt{27} = \pm3\sqrt{3} \)

Which means a = -3sqrt(3) or a = 3sqrt(3)

Ok, thanks, I think I get it.

I hope I can use it appropriately in the future. :)

Just picture all the 2's and 4's with extra dots turning them all into 6's, say.

To all intents and purposes they're the same, same as the 1's, 3's and 5's.

g(-5) equals g(f(x) so f(x) must equal -5. So

3x - 8 = -5.

Solving this gives x = 1. So now g(-5) equals

2 * 25 - 25 - 3 = 22. 

Use a table of trigonometric values to find the angle θ in the right triangle in the following problem. Round to the nearest degree, if necessary.

cos θ = 0.8192         A = 45         H =    ?   

i cant figure out how to set up these equations. in my book i have a table of trigonometric values and it shows 35 degrees for cos 0 = 0.8192

I would use probablility contour mapping to answer this question.

A line segment of length 5 is broken at two random points along its length. What is the probability that the shortest of the three new segments has length longer than 1?

Let the lengths be x,   5-y and  y-x

Note that when you add all these you get a lenght of 5  

so

0

Lets graph that.

The darkest area represents the entire sample space.   How much area does it have?

Area = 0.5*5*5 = 12.5 units squared

Now how much of this sample space will have the shortest side bigger than than 1 unit?

mm

I think it is a lot easier to graph the sample space where one side is less than 1.

These are mutually exclusive and exhaustive events so their probabilities will be add up to 1

The lengths are  x,   5-y and  y-x

When will a length be LESS than 1 unit

\(0

The dark shaded area is where there IS a side that is less then one unit.

SO the triangle in the middle is where there is NO SIDE less than 1 unit.

Area of trianlge in middle = 0.5*2*2 = 2 units squared.

So the probability that the shortest length will be longer than 1 unit is  \(\frac{2}{12.5}= \frac{4}{25}=\frac{16}{100}=16\%\)

Thanks heureka!!!

Thanks guest.

I don't understand the 4!

They do not have to be all the same ???

\(f(x) = \begin{cases} k(x) &\text{if }x>2, \\ 2+(x-2)^2&\text{if }x\leq2. \end{cases}\)

Mmm.

The first thing I notice is that f(2)=2  so this point will be its own inverse (it lies on the line y=x)

That helps becasue it gives me less things I have to think about.

\(\text{This will work if k(x) is the inverse of   } \\f(x)=2+(x-2)^2 \;\; where\;\ x\le 2\)

So I need to find the inverse:

let\(y=2+(x-2)^2    \qquad    x\le 2, \quad y\ge 2\\ \text{I need to make x the subject}\\ y-2=(x-2)^2\\ \pm\sqrt{y-2}=x-2\\ \pm\sqrt{y-2}+2=x\\ \text{But x has to be less then 2 so it must be the neg square root}\\ x=-\sqrt{y-2}+2 \qquad    x\le 2, \quad y\ge 2\\ \text{Now for the inverse swap the x and y values}\\ y=-\sqrt{x-2}+2 \qquad x\ge2, \quad y\le 2 \)

so

\(f(x) = \begin{cases} -\sqrt{x-2}+2 &\text{if }x>2, \\ 2+(x-2)^2&\text{if }x\leq2. \end{cases}\\ \text{will be its own inverse}\)

so

\(g(x)=-\sqrt{x-2}+2 \qquad\text{where}\;\;x>2\)

Here is the graph:

Thank you melody, i understand now!