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Thanks guest,

Yep, Ginger sure got that one wrong :)

Gino is definitely affiliated with rhinos, not hippos :))

how can i solve this? ln(e^x + 1) + x

How could you solve it, there is no equal sign so there is nothing to solve!

Rhinoceros !!!!!!!

noun, plural rhinoceroses (especially collectively) rhinoceros.

1.

any of several large, thick-skinned, perissodactyl mammals of the family Rhinocerotidae, of Africa and India, having one or two upright horns on the snout: all rhinoceroses are endangered.

Hi Max

Your partial fractions, line 3, are the wrong way round, should be  \(\displaystyle \frac{1}{p-2}-\frac{1}{p-1}\), it works out nicely after that.

Tiggsy

Geno, you have the hippo but you are missing the potamus.

 (Using subscripted variables makes it easier to present and keep track of the relations.)

\(\small \text {Let A1 = {pass exam 1}, A2 = {pass exam 2}, A3 = {pass exam 3}, and A = {passes all three exams}. }\\\)

\(\text{Because }\ A^c_2\subset A^c, \; A^c_2\cap Ac=A^c_2 \\ \mathbb{P}(A^c_2) = \mathbb{P}(A^c_2|A^c_1)\; \cdot\; \mathbb{P}(A^c_1 +\mathbb{P}(A^c_2|A_1) \; \cdot\; \mathbb{P}(A_1) \leftarrow \small \text{Apply Total Probability Formula}\\ = (0 \cdot 0.1) + (0.2 \cdot 0.9) = 0.18 \\ \text {Then }\\ \mathbb{P}(A^c_2|A_c) = \dfrac {\mathbb{P}(A^c_2 \cap A^c)} {\mathbb{P}(A^c)}\\ = \dfrac {0.18}{0.496} = 0.363\)

GA

In future please put ALL the question in the body of the question. I am actually not sure what the whole question is!

x= 4 sin t, y=2 csc t, for t in (0,3.14) y=??, for x in ??

It seems to be in radians  \(0

so

\(0 2\\\)

\(y=2csc(t)\\ y=\frac{2}{sin(t)} \\y=\frac{4*2 }{4sin(t)}\\ y=\frac{8 }{x}\\\)

  Here is the graph

Guest, 

We can still get a 'proper' answer with a fraction.

Example:

\(\dfrac{41}{7}\pmod{13}\\ \equiv\dfrac{2}{7}\pmod{13}\text{ because 41 mod 13 = 2}\\ \equiv\dfrac{4}{14}\pmod{13}\\ \equiv \dfrac{4}{1}\pmod {13}\text{ similarly, because 14 mod 13 = 1}\\ \equiv 4 \pmod {13}\\ \equiv 4\)

Or

\(\dfrac{a}{b}\pmod c\\ \text{We first find numbers m and n that } bm = cn+1\\ =\dfrac{am}{bm}\pmod c\\ =\dfrac{am}{cn + 1}\pmod c\\ =am\pmod c\text{ (because (cn + 1) mod c = 1)}\\ \text{So we find the answer.}\)

The probabiltity that she did not pass the second exam.

In order to take the second exam, she must have passes the first.

So, in order to pass the first exam and to fail the second:  0.9 x 0.2  =  0.18.

After failing the second exam, she is not allowed to take the third exam.

I believe that the answer is 0.18.

We can get both of these equations into slope-intercept form, which is....

y  =  mx + b            , where  " m "  is the slope and  " b "  is the y-intercept .

First equation:

y  =  1/3 x - 2           This equation is in slope-intercept form. We can see that m  =  1/3  =  the slope.

Second equation:

-2x + 6y  =  18          Add  2x  to both sides of this equation.

6y  =  2x + 18           Divide through by  6  .

y  =   2/6 x + 3

y  =   1/3 x + 3          Now we can see that  m  =  1/3  =  the slope.

The slopes are both  =  1/3  , and the y-intercepts are different.

This means that the lines are parallel, but they are not the same.

That's a neat way to do " FOIL " ing !!

Sometimes when these exponents get confusing it helps to write out all the letters as many times as the exponent says. Also, we can write out the prime factorization 147...

\(\sqrt{{\color{magenta}147}\,\cdot\,{\color{RedOrange}x^6}\,\cdot\,{\color{teal}y^7}} \\~\\ =\sqrt{{\color{magenta}7\,\cdot\,7\,\cdot\,3}\,\cdot\,{\color{RedOrange}x\,\cdot\,x\,\cdot\,x\,\cdot\,x\,\cdot\,x\,\cdot\,x}\,\cdot\,{\color{teal}y\,\cdot\,y\,\cdot\,y\,\cdot\,y\,\cdot\,y\,\cdot\,y\,\cdot\,y}} \\~\\ =\sqrt{{\color{magenta}7\,\cdot\,7}}\cdot\sqrt{{\color{magenta}3}}\,\cdot\,\sqrt{{\color{RedOrange}x \,\cdot\, x}} \,\cdot\, \sqrt{{\color{RedOrange}x \,\cdot\, x}}\,\cdot\, \sqrt{{\color{RedOrange}x \,\cdot\, {\color{RedOrange}x}}}\,\cdot\,\sqrt{{\color{teal}y\,\cdot\,y}}\,\cdot\,\sqrt{{\color{teal}y\,\cdot\,y}}\,\cdot\,\sqrt{{\color{teal}y\,\cdot\,y}}\,\cdot\,\sqrt{{\color{teal}y}}\)

The square root of  (7 * 7)  is the square root of  7 squared, which is  7  .

The square root of  (x * x)  is  x  .  The square root of  (y * y)  is  y  .

So we can write the original expression as...

\( =7\,\cdot\,\sqrt{3}\,\cdot\,x \,\cdot\, x \,\cdot\,x \,\cdot \, y\,\cdot\,y\,\cdot\,y\,\cdot\,\sqrt{y} \\~\\ =7\,\cdot\,\sqrt3\,\cdot\,x^3\,\cdot\,y^3\,\cdot\,\sqrt{y} \\~\\ =7x^3y^3\sqrt{3y}\)

No idea

There are many factors that go into such an equation: time value of money, interest, etc. etc.

That's a determined child...

In this equation, our goal is to isolate the unknown, x. Let's see how we would go about doing that

There ya go!

No way  to tell........the temperature could have been - 3.2 F everyday, so that it might not have changed at all....

(c-4)(3c+9)

Distribute  the terms in the first set of parentheses over the terms in the second set of parentheses

c (3c + 9)   -  4 (3c + 9)

3c^2 + 9c - 12c - 36    combine like terms

3c^2 - 3c  - 36

Yes, it is possible. In fact, I have the entire answer for you.

\(8239194821^{100}=\)

38768340824283047116920184993265064638556302587932919267287017626749780235349455788600968015625511431133236374272511088887669179317147392235403570848341357292635506626294751155792171416395889300854082342350450792264241428923588462680772019491201775782203898455196723341878450011080592555646946948076749167008032173860304404158904142427370015445333124633440902838859059595743143629991307549967606676248908229220456294819582123391483792591432129184014143747159534734085947546448982777831143851021208773065460769455508529629773589737045363075702583515183968234983485316526056637898322556872587185303656198642805194037193866095810267888147952219872774215271878314010337199920028757633215967346388705402738301953592586870045524269833967864410828357539466559216970703818455121691325786391475544211781337513810830103750086089621097986106065376399926885956704954256747057782360550018568741898368886116309725952333679868551784789206597848287109644486947462735549341823984238772401277572029322725462001

Yeah, that's a bit too big to grasp...

I honestly don't know if the answer above is completely accurate.

When I tried converting it into a fraction I got the final answer of \(\frac{4225771273642583}{5000000000000000}\) by placing the number over 10000000000000000 and dividing the numerator and denominator by its GCF, 2. 

t = 2n – 3....... How long does it take if a customer buys 5 coffees?

t = 2(5) - 3  =   10 - 3   =  7 minutes

a mod b- the remainder of the division a/b.

(P-3)/(P-2) mod P- the remainder of the division ((P-3)/(P-2))/P. P>(P-3)/(P-2),

therefore ((P-3)/(P-2))/P<1, therefore (P-3)/(P-2) mod P=(P-3)/(P-2)-P*0=(P-3)/(P-2)

If f(x) = 3x - 1 then f(-2) means replace the x by -2 in the expression 3x - 1, and calculate the result.

So here we have f(-2) = 3*(-2) - 1 → -6 - 1 → -7

.

When I was in grade 5 I calculated that just using pencil and brain XD

and wrote that number down on my pencil box XDDDDDD

so that's 2^40 = 1,099,511,627,776

\(1^{-1}\cdot 2^{-1} + 2^{-1} \cdot 3^{-1} +... + (p-2)^{-1}\cdot (p-1)^{-1}\\ =\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{(p-1)(p-2)}\\ =(1-\dfrac{1}{2})+(\dfrac{1}{2}-\dfrac{1}{3})+...+(\dfrac{1}{p-1}-\dfrac{1}{p-2})\\ =1-\dfrac{1}{p-2}\\ =\dfrac{p-3}{p-2}\)

So our work is to evaluate

\(\dfrac{p-3}{p-2}\pmod p\).

Don't know how to do that, if I were you, I would try some primes.

So let's try it.

When p = 7,

\(\dfrac{p-3}{p-2}\pmod p\\ =\dfrac{4}{5}\mod 7\\ =\dfrac{12}{15}\mod 7\\ =5\)

When p = 11,

\(\dfrac{p-3}{p-2}\pmod p\\ =\dfrac{8}{9}\pmod {11}\\ =\dfrac{40}{45}\pmod {11}\\ =7\)

When p = 13,

\(\dfrac{p-3}{p-2}\pmod p\\ =\dfrac{10}{11}\pmod {13}\\ =\dfrac{60}{66}\pmod {13}\\ =8\)

When p = 17,

\(\dfrac{14}{15}\pmod{17}\\ =\dfrac{112}{120}\pmod {17}\\ =10\)

Welp can't see a pattern :(