thelizzybeth

Both of these involve using the discriminant \(b^2-4ac\).

1. Since there are no real solutions, \(b^2-4ac < 0\). 

\((-k)^2-4*8*5 < 0\)

\(k^2 < 160\)

The largest integer k is 12.

2. First expand: \(2x^2-3x-35=-43+jx\)

Then combine like terms: \(2x^2-(3+j)x+8=0\)

Since there is one real solution, \(b^2-4ac = 0\)

\((-(3+j))^2-4*2*8=0\)

\(j^2+6j-55=0\)

\((j+11)(j-5)=0\)

So \(j=-11, 5\)

Its x-intercept is (x, 0). 

The slope is \(\frac{0-18}{x-0}=\frac{-18}{x}\). 

We are told this equals -3/7 so \(\frac{-18}{x}=\frac{-3}{7}\).

Solving for x, we find that the x-intercept is (42,0)

There's not a simple way to factorize \(2x^2 - 3x - 1 = 0\), so instead we simplify ^{\(\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}=\frac{(a+b)^2-2ab}{ab}=\frac{(a+b)^2}{ab}-2\)}

You could plug in the values of 2 and 3 to create a system of equations with two variables.

Because Brennan has the lightest bars, and Maya has the heaviest bars, the average weight of Blair's bars must be between their amounts. We can then write an inequality and solve it. Set the total weight as w and the number of bars Blair received as x. 

Average weight = Total weight/Number of bars

So Brennan's average weight = 0.45W/24

And Maya's average weight = 0.26W/13

Since the rest of the bars are given to Blair, she must have (1-0.45-0.26)W as her total weight.

Blair's average weight = 0.29W/x

Now we write our inequality:

\(\frac{0.45W}{24} < \frac{0.29W}{x} < \frac{0.26W}{13}\)

We can divide everyone by W. 

\(\frac{0.45}{24} < \frac{0.29}{x} < \frac{0.26}{13}\)

Then split it into two equations.

\(\frac{0.45}{24} < \frac{0.29}{x}\) and \(\frac{0.29}{x} < \frac{0.26}{13}\)

Simplify by cross-multiplying to get:

\(0.45x<0.29*24\) and \(0.26x>0.29*13\)

Isolate x to get:

x < 15.47 and x > 14.5

14.5 < x < 15.47

The only integer value that satisfies this is 15, which is our final answer.

If a quadratic has no real roots, the discriminant \(b^2-4ac\), where a=1, h+2=b and h+5=c, is less than 0. 

So let's plug in our values.

\((h+2)^2-4(h+5)<0\).

Expand:

\(h^2+4h+4-4h-20<0\)

Simplify by combining like terms:

\(h^2-16<0\)

This can be factorized into:

\((h+4)(h-4)<0\)

This can only happen when one is positive, and one term is negative. h+4 is obviously greater than h-4, so h+4 must be the positive one.

\(h+4>0\)  and  \(h-4<0\)

\(h>-4\)  and  \(h<4\).

Combine to get \(-4 < h < 4\)

Since h is the greatest positive integer, the answer is 3. 

Using the midpoint formula, we find that the midpoint of the first segment is

\((\frac{2+0}{2}, \frac{4-2}{2})\)=\((\frac{2}{2}, \frac{2}{2})\)=\((1, 1)\)

The midpoint of the second segment is 

\((\frac{5+1}{2}, \frac{1+5}{2})\)=\((\frac{6}{2}, \frac{6}{2})\)=\((3, 3)\).

Using the slope formula, we find that the slope of the line with points \((1, 1)\) and \((3, 3)\) is:

\(\frac{3-1}{3-1}=\frac{2}{2}=1\)

Let's say there are x teams. 

Each team captain shakes hands x-1 times (everyone except themselves). There are x team captains so there are x(x-1) handshakes. However, we have to divide by two because for one handshake, we counted twice- once for each person shaking hands. So in all, there are x(x-1)/2 handshakes.

This equals 595 so x(x-1)/2=595

x(x-1)=595*2

\(x^2-x-595*2=0\)

Prime factorizing 595*2 = 2*5*7*17*= 34*35

So we can factorize as (x-35)(x+34)=0

We get two solutions, 35 & -34. Since the number must be positive, the answer is 35. 

Set our number as x and turn the words into math:

\(\frac{1}{4}x=3*19\)

\(x=3*19*4\)

\(\frac{1}{6}x=\frac{3*19*4}{6}\)

\(\frac{1}{6}x=38\)

First, think of Sam, Rebecca, and Diane as a "block" or one person. So there are now 6 people to consider (five other people and the "block"). There are \(\frac{6!}{6}\) ways these 6 people can sit around a circular table. Six options for the first seat, five options for the second, etc. We divide by 6 because for one seating arrangement, we can rotate the table six times. 

But we can't forget about the seating arrangments within the block. Rebecca can either be on the right side of Sam, and Diane is then on the left, or vice versa. So we have to multiply by two.

Solve for this to get our answer \(\frac{6!}{6}*2\) = 240.