guestperson

(b) Suppose we have \(n\) people and we want to make a committee of arbitary size. The amount of ways we can choose a committee of 0 people is \(\dbinom{n}{0}\). The amount of ways we can choose a committee of 1 person is \(\dbinom{n}{1}\). In general, the amount of ways we can choose a committee of k people is \(\dbinom{n}{k}\). Then, the total would be 

                                                                                        \(\dbinom{n}{0} + \dbinom{n}{1} + \dbinom{n}{2} + \cdots + \dbinom{n}{n}\) 

ways. However, notice that this is also equivalent to choosing whether or not a person should be on the committee. Each person has 2 choices - be on the committee or don't. Then, there are \(2^n\) total ways. Conclusively, \(\dbinom{n}{0} + \dbinom{n}{1} + \dbinom{n}{2} + \cdots + \dbinom{n}{n}\)=\(2^n\).

We can do this by using some simple arithmetic: 

\(13 \frac{1}{8} = 13 + \frac{1}{8}\).

This lets us separate the sum and focus on finding \(\frac{1}{8}\). \(\frac{1}{8} = 0.125\) from simple long division, so we have 

\(13 + \frac{1}{8} = 13+ 0.125 = 13.125\). 

Guest, I assume you mean how many ways you can put 7 textbooks on the bookshelf . 

If this is true, we can use some simple counting to solve this. We know that this bookshelf has to be of the form

_ _ _ M _ _ _ 

if M is a math textbook. So we simply need to find ways to arrange the other textbooks. We know all of the textbooks are different from the problem, so we do not have to worry about distinguishability. Thus, there are \(6! = 720\) ways of arrangement. 

Because 493 is neither a perfect square nor is divisible by any perfect squares, it cannot be simplified further from \(\sqrt{493}\). The best we can do is use some rounding to get 

\(\sqrt{493} \approx 22.203603...\)

Alternative Solution: 

Notice that \((x+y)(x-y) = x^2-y^2\), so simply multiplying the two together gives: 

\((x+y)(x-y) = x^2-y^2 = \frac{15}{7} \cdot \frac{7}{9} = \frac{15}{9} = \boxed{\frac{5}{3}}\)

The number N has a remainder of 2 when divided by 3 and 4, so the number will have a remainder of 2 when divided by \(3\cdot 4 = 12\). Thus, N is of the form \(N = 12x+2\). This easily shows that the remainder when N is divided by 12 is \(\boxed{2}\) . 

Hope this helped :P

There are 40 voters (50-10 =40since. we have to exclude the candidates). Each voter has 11 choices, since they can choose from 10 candidates or choose none (abstaining). So, the total amount of different possible vote totals is \(\boxed{40\cdot11 = 440}\). 

Hope this helped 

I'm assuming you're solving \((3x+12)(x+4) = \frac{4x^2+8x+8x^2}{2x}\). With this, start by simplifying the RHS (right hand side) to \(2x+4+4x\), which is equal to \(6x+4\). 

Now, expand the LHS (left hand side) to get that \(3x^2+24x+48 = 6x+4\). Shift the stuff on the RHS to the LHS to get \(3x^2+18x+42=0\). Using the quadratic equation, the answers are \(x=-3 + \sqrt{5}i\) and \(x = -3-\sqrt{5}i\) (complex roots). 

Hope this helped 

Using modular arithmetic, we can create three "equations." 

\(n\equiv 5 \pmod{9} \)

\(n\equiv 2 \pmod{7}\)

\(n\equiv4\pmod{5} \)

from the last equation, we get that \(n = 5a+4 \) for some integer a. From the middle one, we get \(n = 7b+2\) for some integer b. Set these equal to each modulo 5 to get that\(5a + 4 \equiv 7b+2 \pmod{5}\). you get that \(2b\equiv 2\pmod{5}\), so \(b \equiv 1 \pmod{5}\). So, \(b = 5c+1\) for some integer c and the bottom system simplifies to 7(5c+1) + 2 = 35c+9, so \(n\equiv 9\pmod{35} \). 

We do the same with this and the top equation we left out before to get that \(n = 315e + 149\) for some integer e. e=3 minimizes the four digit integer, and we get n = 1094. 

hope this helped 

Start with the bread. There are 5 types of bread and we choose 2 from them. \(\dbinom{5}{2} = \dfrac{5!}{3!\cdot2!} = 10\) different combinations. There are 7 different ingredients and the order does not matter, so we get \(7! = 5040\) ways to distribute the ingredients. The final answer is \(10 \cdot 5040 = 50400\) 

hope this helped 

Since it has zeroes at \(x = 2\) and \(x = 4\), the quadratic can be written in the form \(y = a(x-2)(x-4)\). You can substitute the point \((3,6)\) to get that \(6=a(3-2)(3-4)\), so \(a=-6\). Thus, we get the answer of \(-6(x-2)(x-4) = -6(x^2-6x+8) = -6x^2 +36x-48\). 

hope this helped 

Let \(t=\) ticket price. There are 4 tickets, which means \(4t\) is the price without tax. With tax, this is multiplied by \(1.095\). So, we get \(1.095(4t) = 72.27\). Divide both sides by \(1.095 \cdot 4\) to get that each ticket is worth \($16.50\). 

hope this helped 

Graphing it on desmos, the interval from (-4,0) is entirely skipped, so the range is. \((-\infty, -4] \cup[0,\infty)\)

hope this helped 

The axis of symmetry is equal to the average of the roots. When we factor \(x^2-2x-3\), we get \((x+1)(x-3)\), so the roots are -1 and 3 as shown in the graph. \(\frac{-1+3}{2} = 1\), so the axis of symmetry is \(x=1\). 

Hope this helped 

\((-\infty,-1)\)