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Figures, I knew the actual equation but I thought that was only for - and not +. Thanks for correcting me though!

Yes the number are 

a=2

b=12

c=?

That's not quite right RP!  The general expression is a² - b² = (a + b)(a - b)

Here we have a = y, but b = 7i, where i is the square root of -1.

So y² + 49 → y² - (7i)² → (y + 7i)(y - 7i)     (Note that i² = -1)

Either that or you can't factor it at all

I believe this is how you do it.

Use the formula to factor: a^2+b^2=(a+b)(a-b)

y^2 is already a perfect square.

49=7*7=7^2 and that's your other perfect square.

Now your expression is y^2+7^2=(y-7)(y+7)

Are there any numbers to plug in or an image of some sort? 

Also 1c=positive

2c=negative

And I dont know how to get b because I dont understand

Is 1b even

and 2b odd

2a would be \(x\rightarrow\infty,f(x)=-\infty\)

\(x\rightarrow-\infty,f(x)=-\infty\) am i right

Hi Rainbow Panda,

1a)      \(\displaystyle\lim_{x\rightarrow \infty} f(x) = \infty\\ \\~\\ \displaystyle\lim_{x\rightarrow -\infty} f(x) =- \infty\\\)

 You can do 2a yourself

 b)

The x axis is the axis of symmetry of all even graphs   That is   f(-x) = f(x)

Odd functions  have 180 degree rotational symmetry about (0,0)    That mean    f(-x) = -f(x)

Are either of these even or odd?

c)  If it finishes in the top right corner where y is positive then the leading coefficient is  +

If it finishes in the bottom right corner where y is negative then the leading coefficient is  -

The leading coefficient is the number in front of the highest power of x.

So now you should be able to answer the rest yourself ;)  

1   is that 'SPECIFIC gravity'    ???

2    Mass is independent of gravity..... WEIGHT is affected by gravity.

          18 oz = 510.29 gm

I still don't undetstand how you got that formula, but it's wrong.

The number of intersections inside a polygonal can't be larger than the number of diagonals squared and the formula for that number is a polynomial P(n) while your formula is an exponential formula.

I had mathematica draw the lines for 5, 6, and 7, points.

Counted intersections by eye for each.  Worked out the pattern.

Are you sure you have written the equations correctly?  g(x) must be of order 2, so its highest order term is ax². This means we must have 5a² = -8   (because 5a²x⁴ must equal -8x⁴).  This would make 'a' an imaginary number!

^{There was a flat containing boxes of apples having a total weight of 100 kg. An analysis showed that the apples were 99% moisture, by weight. After two days in the sun, a second analysis showed that the moisture content of the apples was only 98%, by weight. What was the total weight of the apples after 2 days, in kg?}

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My statistics professor often posed questions like this to reinforce the importance of paying attention to how a question presents data and correctly interpreting data that results from analysis.

Solution:

\(\text {The first analysis shows the apples are 99% water. The weight of the water is then}\\ \left(0.99\cdot 100\right) = 99 ~ kg\\ \text {Let x be the weight of the water lost after exposure to the sun. }\\ \left(0.99 \cdot 100-0.98(100-x)\right)=x\\ \begin{aligned}99-(98-0.98x)&=x\\99-98+0.98x&=x\\1+0.98x&=x\end{aligned}\\ \begin{aligned}1+0.98x-0.98x&=x-0.98x\\1&=0.02x\\x&=50\\\end{aligned}\\ \text { }\\ 100-x=100-50=50 ~ kg \leftarrow \text { The total weight of the apples after 2 days}\\\)

The correct solution appears paradoxical. It’s not, but it is counter-intuitive.

I got 126 intersections, how did you get that formula?

Find the constant term in the expansion of $$\left(10x^3-\frac{1}{2x^2}\right)^{5}$$

\(\begin{array}{|rcll|} \hline && \left(10x^3-\dfrac{1}{2x^2}\right)^{5} \\\\ &=& \left(\dfrac{10x^3\cdot 2x^2-1}{2x^2}\right)^{5} \\\\ &=& \left(\dfrac{20x^5-1}{2x^2}\right)^{5} \\\\ &=&\dfrac{1}{\left(2x^2 \right)^5} \left( 20x^5-1 \right)^{5} \\\\ &=&\dfrac{1}{ 2^5x^{10} } \left( 20x^5-1 \right)^{5} \\\\ &=&\dfrac{1}{ 32x^{10} } \left( 20x^5-1 \right)^{5} \\\\ &=&\dfrac{1}{ 32x^{10} } \left[ \dbinom{5}{0}\left(20x^5 \right)^5 -\dbinom{5}{1}\left(20x^5 \right)^4 +\dbinom{5}{2}\left(20x^5 \right)^3 \\ -\dbinom{5}{3}\left(20x^5 \right)^2 +\dbinom{5}{4}\left(20x^5 \right) +\dbinom{5}{5}(-1)^5 \right] \\\\ &=&\dfrac{1}{ 32x^{10} } \left[ \dbinom{5}{0} 20^5x^{25} -\dbinom{5}{1} 20^4x^{20} +\dbinom{5}{2} 20^3x^{15} \\ { \color{red}-\dbinom{5}{3} 20^2x^{10} } +\dbinom{5}{4} 20x^5 -\dbinom{5}{5} \right] \\ \hline \end{array}\)

Constant term:

\(\begin{array}{|rcll|} \hline && \dfrac{1}{ 32x^{10} }\left[ { \color{red}-\dbinom{5}{3} 20^2x^{10} } \right] \\\\ &=& -\dfrac{1}{ 32x^{10} } \dbinom{5}{3} 20^2x^{10} \\\\ &=& -\dfrac{20^2}{ 32 } \dbinom{5}{3} \quad & | \quad \dbinom{5}{3} = \dbinom{5}{5-3}=\dbinom{5}{2}=\dfrac{5}{2}\cdot \dfrac{4}{1} = 10 \\\\ &=& -\dfrac{20^2\cdot 10}{ 32 } \\\\ &=& -\dfrac{4000}{ 32 } \\\\ &\mathbf{=}& \mathbf{-125} \\ \hline \end{array}\)

You are at a right place  Happy weekend plus sunday, the funday and saturday night is really stunning.

Find all numbers r for which the system of congruences:

\(\begin{align*} x &\equiv r \pmod{6}, \\ x &\equiv 9 \pmod{20}, \\ x &\equiv 4 \pmod{45} \end{align*}\)

has a solution.

\(\begin{array}{|rclll|} \hline x &\equiv& 9 \pmod{20} \\ x &\equiv& 4 \pmod{45} \\ \text{or} \\ x &=& 9 + 20n \quad | \quad n \in Z \\ x &=& 4 + 45m \quad | \quad m \in Z \\\\ && & x = 9 + 20n = 4 + 45m \\\\ && & 20n = 45m - 5 \quad & | \quad : 5 \\\\ && & 4n = 9m - 1 \\\\ && & \mathbf{ n = \dfrac{9m-1}{4} }\\\\ && & n = \dfrac{8m+m-1}{4} \\\\ && & n = 2m+\underbrace{\dfrac{m-1}{4}}_{=a} \\\\ && && a = \dfrac{m-1}{4} \\\\ && && 4a = m-1 \\\\ && & & \mathbf{ m = 4a + 1 } \\\\ && & \mathbf{ n = \dfrac{9(4a + 1 )-1}{4} }\\\\ && & n = 9a + \dfrac{8}{4} \\\\ && & \mathbf{ n = 9a + 2 }\\\\ \mathbf{x} &\mathbf{=}& \mathbf{9 + 20(9a + 2)} \\\\ x & = & 9+180a+40 \\\\ \mathbf{ x } &\mathbf{ = } & \mathbf{49+180a} \quad | \quad n \in Z \\ \hline \end{array} \)

\(\begin{align*} x &\pmod{20}:& &49\pmod{20}+(180a)\pmod{20} \\ && &= 9 + 0 \\ && &= 9\ \checkmark \\\\ x &\pmod{45}:& &49\pmod{45}+(180a)\pmod{45} \\ && &= 4 + 0 \\ && &= 4\ \checkmark \\\\ r &\equiv x \pmod{6}:&&r = 49\pmod{6}+(180a)\pmod{6} \\ && &r = 1 + 0 \\ && & \mathbf{r = 1} \\ \end{align*}\)

What is 3/2 to the 3rd power?

\(\begin{array}{|rcll|} \hline && \mathbf{ \left(\dfrac32 \right)^3 } \\\\ &=& \left(\dfrac32 \right)\cdot \left(\dfrac32 \right)\cdot \left(\dfrac32 \right) \\\\ &=& \dfrac{3\cdot 3\cdot 3 } {2\cdot 2 \cdot 2} \\\\ &=& \dfrac{27} {8} \\\\ &\mathbf{=}& \mathbf{3.375} \\ \hline \end{array}\)

Suppose person 1 is telling the truth.

Then person #2 says that at least 2/3 of person 1 has told a lie which I will take to mean he says person 1 lied.

But we know that person 1 told the truth so this means that person #2 lied.

But person #1 truthfully said that none of them lied so we have a contradiction.

Thus person 1 must have lied.

If you work through the responses and the count of liars you see that 2/3 of the people are liars i.e. knaves.

\(\dfrac 2 3 \cdot 2019 = 673\)

One of the laws of exponents state that:
\((\frac{a}{b})^x = \frac{a^x}{b^x}\)
With that in mind, 

\((\frac{3}{2})^3 = \frac{3^3}{2^3} = \frac{27}{8}\)