NotThatSmart

There are 4 children, which means there are 4! = 4 x 3 x 2 x 1 = 24 ways to seat the children.

There are also 4 adults, which means there are, again, 4! = 4 x 3 x 2 x 1 = 24 ways to seat the adults.

Now, we just have to do 24*24 = 576 since each arrangement of children can be met by 24 ways to arrange the adults.  

There are 576 ways to do this!

Thanks! :)

In order to find all of ordered pairs that work, we just have to write an equation. 

If the two graphs intersect, then we have $x^3 + x^2 - 3x + 5 = x^3+2x^2$. 

Combing all like terms and moving all terms to one side, we get $x^2+3x-5 = 0$. 

Using the quadratic equation, we get 

$x=\frac{\sqrt{29}-3}{2} \approx 1.193\\ x=\frac{-\sqrt{29}-3}{2} \approx -4.193$

Plugging these two values back into our equations from earlier, we get

$y \approx 4.541 \\ y\approx -38.541$

So our two orders pairs approximately are

$(1.193, 4.541), (-4.193, -38.541)$

Thanks! :)

So first off, let's calculate the distance Will fovers before Grace even begins to row. 

We have $(45 min)(50 m/min) = 2250 m$. 

When Grace starts to row, the two have to cover $(2800 m) – (2250 m) = 550 m$

Since they are rowing at eachother at the same time, if we let t be the time it takes, we have $(50)(t) + (30)(t) = 550$. 

Solving this, we get

 $80t = 550 \\ t = 550/80 = 6.875$

This rounds to approximately 6 minutes and 52 seconds. 

This means in addition to the 45 minutes Will takes at the beggining, they will take an additional 6 minutes and 52 seconds to meet. 

Now, we some math to find that they meet at $2:51:52$ P.M. 

Thanks! :)

First, let's multiply g(x) by f(x). I won't show all the steps cuz I'm a bit lazy....

But make sure use the Distributive property to expand $\left(\:-x^2+\:8x\:-\:5\right)\left(\:x^3\:-\:11x^2\:+\:2x\right)$. 

In the end, we get

$-x^5+19x^4-95x^3+71x^2-10x$

The leading term is the term that is first, which is ofcourse, $-x^5$

The leading coefficient is the coefficient of the leading term, which is $-1$

The degree is the highest raised power, which is $5$

The constant term is the term with no x values, which there is none in this case, so $0$

The coefficient of x^2 is clearly $71$

I hope I answered all of your questions!

Thanks! :)

Let the money of the bill be x. 

(a) From the problem, we know that $\frac{x/3 + 9}{x/3 + (x/3-9)} = \frac{5}{6}$

We just have to solve for x now. 

$\frac{x+27}{2x-27}=\frac{5}{6}$

$6x+162=10x-135 \\ 4x=297\\x=74.26$

The total bill was $74.25

(b) We just have to work backwards now. 

First off, we know at first, Calvin had to pay $74.25/3 = 24.75$ dollars. 

However, he then had 9 dollars paid for him, so we have $24.75-9=15.75$. 

Calvin had to pay Bala $15.75 

Thanks! :)

First, let's write this in factored form. 

A quadratic whose roots are 5 and -5 can be written in the form (x-5)(x+5). 

Expanding, we have $(x-5)(x+5) = x^2-25$. 

This means that the first blank, the coefficient for x, is 0. 

The other blank, the constant, is -25. 

We have a final answer of $x^2+0x+25$ or just $x^2-25$

Thanks! :)

I'm not exactly sure if I'm doing correctly, but I'll try my best. 

First off, let's note that $a^x \cdot a^y = a^{x+y}$. 

Using this logic, we would have $q^{\frac{5}{3}}p^{\frac{5}{4}}r^{\frac{31}{12}}$. 

In order to not have fractional exponents, we would have to put these into radical form. 

Note that $a^{x/y} = \sqrt[y]{a^x}$.

We would get $\sqrt[3]{q^5} \cdot \sqrt[4]{p^5} \cdot \sqrt[12]{r^{31}}$.

I hope I answered your question!

Thanks! :) 

3. As I said in question #2, we have $(x – h)^2 + (y – k)^2 = r^2$ where r is the radius. 

This means we have $r^2 =169$, which just gives us $r = \sqrt{169} = 13$. So the radius is 13. 

Note, the radius can't be -13 because the length or radius can't be negative!

Thanks! :)