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Thank you MathProblemSolver101

Wow @CPhill

That solution was brilliant!

The values of $x$ and $y$ are always positive, and $x^2$ and $y$ vary inversely. If $y$ is 10 when $x$ is 2, then find $x$ when $y$ is 4000.

$y = \frac{k}{x^2}, k \in \mathbb{R}$ 

$k = x^2 \cdot y = 2^2 \cdot 10 = 40$

$y = \frac{40}{x^2}$

$4000 = \frac{40}{x^2}$

$100 = \frac{1}{x^2}$

$100x^2 = 1$

$x^2 = \frac{1}{100}$

$x = \frac{1}{10}$

x +  y =  3      

xy   = 1   ⇒  y  = 1/x

So

x +  1/x  = 3        multiply through  by   x

x^2  + 1  =   3x

x^2  - 3x    =   -1             complete the  square on x

x^2  - 3x  + 9/4  =  -1  +  9/4

(x - 3/2)^2   =  5/4         take both roots

x - 3/2  =  -sqrt (5/4)           or       x   - 3/2  = sqrt (5/4)

x  - 3/2  =   -sqrt (5) / 2                    or       x  - 3/2  =  sqrt (5) /2

x =    (3 -sqrt (5)) / 2                                   x =  (3 + sqrt (5) ) / 2

a   =  3         b = 1        c = 5      d = 2

Sum =   11

Hint: Evaluate each of $\binom{n}{k-1}, \binom{n}{k}, \binom{n}{k+1}$ (which can be found in heureka's answer if you're stuck.)

Hint 2: You know $\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n - k + 1}{k},$ you have $\binom{n}{k} = \binom{n}{k-1} \left(\frac{n - k + 1}{k}\right)$ and $\binom{n}{k-1} = \frac{k \binom{n}{k}}{n-k + 1}$. Evaluate $\binom{n}{k+1}$ in terms of $\binom{n}{k}$

a)

Notice that adding 2 to a number changes its remainder when divided by 3 by -1. That means the possible remainders (n, n+2, n+4) can leave when divided by 3 is: (-1, 1, 0), (0, -1, 1), and (1, 0, -1), so no matter what, at least one of (n, n+2, n+4) is divisible by 3. That means the only possibility is that n is 3, so the only triplet of primes in that form is (3, 5, 7)

b)

Using a similar logic as above, we can tell that in a group of 5 evenly spaced numbers where the difference is not a multiple of 5, exactly one of them is divisible by 5. 

That means you have to set the first term equal to five and add a non-multiple of 5 4 times until you get one of them that makes all the numbers prime. can you finish it?

Ah, yes.....I see my mistake....I read it as  the  "same square".....thanks for the correction  !!!!

For  the second one......here's a similar problem solved  by heureka.....see if it might help.....

https://web2.0calc.com/questions/binomial-theorem-question-pls-help

Nice answer :)

alternate solution to 7.1.3 without knowing euler's formula(as it is only learned in calculus i believe):

\(\frac{w}{z}\\ = \frac{r}{s}\cdot\frac{\cos(\alpha)+i\sin(\alpha)}{\cos{\beta}+i\sin{\beta}}\\ = \frac{r}{s}\cdot\frac{(\cos(\alpha)+i\sin(\alpha))(\cos{\beta}-i\sin{\beta})} {(\cos{\beta}+i\sin{\beta})(\cos{\beta}-i\sin{\beta})}\\ = \frac{r}{s}\cdot\frac{(\cos(\alpha)+i\sin(\alpha))(\cos{\beta}-i\sin{\beta})}{\cos^2{\beta}+\sin^2{\beta}}\\ = \frac{r}{s}\cdot\frac{\cos{a}\cos{b}-i\sin{b}\cos{a}+i\sin{a}\cos{b}+\sin{\beta}\sin{\alpha}}{\cos^2{\beta}+\sin^2{\beta}}\)

by the Pythagorean identity and angle difference rules, this reduces to:

\(\frac{r}{s}\cdot(\cos(\alpha-\beta)-i\sin(\alpha-\beta))\\=\boxed{\frac{r}{s}cis(\alpha-\beta)}\)

Also, for 7.1.4, try converting them into polar form, because it will make it easier to multiply (in your head, of course). 

you're right, counting each solution would take forever.

Notice that the prime factorization of 720 is \(2^4\cdot3^2\cdot5\). Obviously, both numbers have to be factors of 720, meaning that the powers of the primes in their prime factorization have to be smaller or equal to the powers of the primes of the prime factorization of 720. Also, note that, for each prime, there must be at least one of the 2 numbers, m and n, that have the power of that prime in its prime factorization equal to that of 720; the other number can have the power of that prime in its prime factorization less than equal to that of 720.

Let n be expressed as: \(2^a\cdot3^b\cdot5^c\), and let m be expressed as \(2^x\cdot3^y\cdot5^z\).  Let's do some casework to figure out the cases:

For a and x, these are the ways we can set them equal to:

a=4, x=0

x=4, a=0

a=4, x=1

x=4, a=1

x=4, a=2

a=4, x=2

x=4, a=3

a=4, x=3

a=4, x=4

which is 9 ways in total.

If you were to do a similar process with b, y and c, z, you would get that the number of ways we can set them equal to is 3*2-1=5 and 2*2-1=3, respectively, so the answer is 3*5*9=135

(sorry if this seems confusing; I am bad at explaining things)

5km^2 = 5*10^10 cm^2   (1km = 10^5cm)

Scale this down by 20000^2 for the map (map scales are usually linear values)

Area on map = 5*10^10/(2*10^4)^2 cm^2 = 5*10^10/(4*10^8) cm^2 = 125 cm^2

Here is 7.1.3

This should also help with 7.1.5 if you were struggling to understand the meaning of cis.

For 7.1.4 multiply out the bracketed terms (using foil), then collect real and imaginary terms together, remembering that i^2 = -1.

maybe because someone posted something before you but they deleted it?

I'm not entirely sure about my answers, as I'm not the best with cylindrical coordinates, but I'll try my best

First one:

The distance can be represented, by the Pythagorean theorem, as:

\(\sqrt{(p\sin(\phi)\cos(\theta))^2+(p\sin(\phi)\sin(\theta))^2+(p\cos(\phi))^2}\\ \sqrt{p^2(\sin(\phi)\cos(\theta))^2+p^2(\sin(\phi)\sin(\theta))^2+p^2(\cos(\phi))^2}\\ =p\sqrt{(\sin(\phi)\cos(\theta))^2+(\sin(\phi)\sin(\theta))^2+(\cos(\phi))^2}\)

We just need to prove that the expression under the radical is always one (also, note that \(\sin^2(x)+\cos^2(x)=1\) for any x by the Pythagorean identity):

\((\sin(\phi)\cos(\theta))^2+(\sin(\phi)\sin(\theta))^2+(\cos(\phi))^2\\ =\sin^2(\phi)\cos^2(\theta)+\sin(\phi)^2\sin^2(\theta)+\cos^2(\phi)\\ =\sin^2(\phi)(\cos^2(\theta)+\sin^2(\theta))+\cos^2(\phi)\\ =\sin^2(\phi)+\cos^2(\theta)\\ =1\)

Because of that, the expression reduces to \(p\), which is our desired answer. (obviously if p<0, the distance would be negative, which is impossible)

Second one:

(a)  \((r, \theta, -z)\), because the z-axis determines the height, and reflecting through the xy-plane is basically inverting the height.

(b) \((r, \theta\pm \pi, z)\), because adding/subtracting pi radians makes no difference as they both reflect \((r, \theta)\) across the origin.

(c) I think it would be a combination of the previous two, namely \((r, \theta \pm \pi, -z)\)

also, I'm kinda confused on why my post is shown as being the second post, not the first.

Let triangle ABC have side lengths AB=13, AC=14, and BC=15.
There are two circles located inside angle BAC
which are tangent to rays AB, AC, and segment BC.
Compute the distance between the centers of these two circles.

A carpenter chisels a square hole of side 2 in. through a round post of radius 2 in. The axis of the hole
intersecting that of the post at right angles. Find the volume of the wood cut out.

Hello Guest!

The face of the side of the body is made up of two segments of a circle and four right triangles.

\(A_{ segment}=\frac{\pi r^2}{6}=\frac{\pi \cdot2^2in^2}{6}=\color{blue}\frac{2}{3}\pi\ in^2\)

\(A_{triangle }=\frac{2\cdot sin(30°)\cdot 1 }{2}=\color{blue}\frac{1}{2}in^2\)

\(A_{body}=2\cdot A_{ segment}+4\cdot A_{triangle }=2\cdot \frac{2}{3}\pi\ in^2+4\cdot \frac{1}{2}in^2=\color{blue}6.189\ in^2\)

\(V_{body}=A_{body}\cdot 2\ in\\ V_{body}=6.189\ in^2\cdot 2\ in\\ \color{blue}V _{body}=12.378\ in^3\)

The carpenter cut 12.378 cubic inches of wood from the log.

  !

Each solution to x^2 + 5x + 18 = 0 can be written in the form a + bi where a and b are real numbers. What is a^2 + b^2?

Hello Guest!

\(x^2 + 5x + 18 = 0\\ x=-2.5\pm \sqrt{6.25-18}\\ a+bi=-2.5\pm \sqrt{11.75}\cdot i\)

\(a^2+b^2=6,25-11.75\)

\(a^2+b^2=-5.5\)

  !

\(\begin{align*} \frac{100\%}{120\%} \cdot 2.52 &= \frac{5}{6} \cdot 2.52 \\ &= 5 \cdot 0.42 \\ &= \boxed{2.1} \end{align*}\)

Chris, I believe that one of us has misread the question:  

What is the probability that both of the dice shows a square number?  

I could multiply the probabilities, but I believe a visual would be more explanatory to the OP guest.  

Here are the possible rolls, with the rolls that solve the problem highlighted.  

1,1   1,2   1,3   1,4   1,5   1,6  

2,1   2,2   2,3   2,4   2,5   2,6  

3,1   3,2   3,3   3,4   3,5   3,6  

4,1   4,2   4,3   4,4   4,5   4,6  

5,1   5,2   5,3   5,4   5,5   5,6  

6,1   6,2   6,3   6,4   6,5   6,6

The probability is shown to be 4/36 = 1/9.

Answer: \(2.1\) \(kilograms\)

Solution:

120% is the same as 6/5, so the cat has 6/5 the mass in June than what it did in March. Set the cat's mass in March to x to create the equation:

\(\frac65x=2.52\)

Multiplying both sides by 5/6 gives x = 2.1. So, Deb's cat's mass was \(2.1\) \(kilograms\) in March.

Hey there, Guest!

So...

Travelling at 8 Km at 80 km/hr takes 1/10 hours while travelling at \(X\) km at 100 km/hr takes \(\frac{x}{100}\) hours, and travelling at \(Y\) km at 60 km/hr takes \(\frac{y}{60}\) hours; in one direction, the sum of these is 2 hours, so \(\frac{1}{10}\) + \(\frac{x}{100}\) + \(\frac{y}{60}\) = 2. In the other direction, however, whatever was uphill becomes downhill, etc.; so the roles of x and y change, and we get the equation:

1/10 + x/60 +y/100 =3.

That equation can be simplified to get:

6x + 10y = 1140

&

6y + 10x = 1740

Therefore the answer is x=165 km, and y = 15 km.

So the total distance is 165 + 15 + 8 = 188 km.

Hope this helped! :)

( ﾟдﾟ)つ Bye

Hey there, Guest!

Alright, so let's write the individual worths of these variables:

a = a

b = a + 2

c = a + 4

So...

b^2 - a^2 = 348

(b+a)(b-a) = 348

(a+2+a)(a+2-a) = 348

Therefore a = 2.

Let's insert that into both of the equations for b and c.

b =a + 2

b = 2 + 2

b = 3

c = a +4

c = 2 + 4

c = 6

And here we are, back at the original equation, except now we know that:

a = 2

b = 3

c = 6

Let's insert that information into the question:

c^2 - b^2

6^2 - 3^2

36 - 9

= 27

Therefore the answer is 27.

Hope this helped! :)

( ﾟдﾟ)つ Bye

Try using a Taylor series expansion.

Since  f(0)  = 4.....then....

c =  4

f(1) =  -10  implies  that

a  +  b   +  4  =  -10

a + b  =   -14

But  a  = `1

So b =  -15

f (x)  = x^2  - 15x   + 4