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\(x = \frac{m \pm \sqrt{n}}{p}\)

2x^2 - 5x - 4 = -x^2 -7x + 6       rearrange as

3x^2 + 2x - 10  =  0 

The  discriminant =  

(2)^2  - 4(3)(-10)  =  

4 + 120  =  

124

sqrt (124)  =sqrt (31 * 4)  =  2 sqrt (31)

So

x =  [ -2  ± 2 sqrt (31)  ]  / ( 2 * 3)  =    [ -1  ± sqrt (31)] / 3

n = 31

Draw SM parallel  to TP  ....let M lie on QP

Draw TL perpendicular to SM......let L lie on SM

Draw QN perpendicular to TP....let QN intersect  SM  at O  and TP at N

In triangle  TLS ....since SM is  parallel to  TP....then  angle TLS = 90

And since TL is perpendicular to SM...angle LTP   =  90  so angle LTS =  120 - 90  = 30 

And angle TSL = 60

So triangle TSL =  30-60-90.... TS = 2 ,  LS  = 1  and TL  =sqrt (3)

And in triangle QOS,  angle QOS = 90

And angle OSQ = 120 -angle LST = 120 -60 = 60 

So triangle SOQ  also is 30- 60- 90

And  QS = 4  so OS= 2 and QO = 2sqrt 3

And  OS - LS  =  2 - 1   = 1  = LO =  TN

And NP =  PT -  TN =  5 - 1  =  4

So triangle QNP is a right triangle with  leg  NP = 4  and  leg QN = TL + QO =  3 sqrt 3  =  sqrt (27)

So  PQ  = sqrt [ 4^2 + (sqrt (27))^2  ]  = sqrt [ 16 + 27 ]  =  sqrt (43)

We can think of the bride and groom as a single unit, so there are 6 people to line up. We can do this in 6! ways. However, the bride and groom can switch places, so we need to divide by 2. Therefore, the total number of ways to line up is 6!/2​=360​.

Another way to solve this problem is to think of the bride and groom as being in the same position. Then there are 5 people to line up, which can be done in 5! ways. But since the bride and groom can switch places, we need to multiply by 2. This gives us the same answer of 6!/2​=360.

thanks!

Let the price of rug 10 years ago ==P

"Appreciate" - means it goes up in value by 5% per year

P ==12,000 / 1.05^10

P ==12,000 / 1.628895

P ==$7,366.96 - price of the rug 10 years ago

Let the point on the x-axis be (x,0). Since this point is equidistant from A and B, we have

[\sqrt{(x + 2)^2 + 0^2} = \sqrt{(x - 0)^2 + 4^2}]or (x+2)2=x2+16. Solving, we find x=−1​.

Since the circumcircles of triangles ABX and ACX have the same radius, then triangle AXB is similar to triangle ACX.

Let r be the radius of the circumcircles of triangles AXB and ACX. Then the circumradius of triangle ABC is also r.

By Power of a Point,

\begin{align*} (AX)^2 &= r^2 + 10^2 \ (CX)^2 &= r^2 + 4^2 \ (BX)^2 &= r^2 + (AX - CX)^2 \ &= r^2 + 13^2 - 8^2 \ &= r^2 + 144 - 64 \ &= r^2 + 80. \end{align*}Adding these equations, we get

[279 = 3r^2.]Then r=93​.

The area of triangle ABC is

[\frac{1}{2} \cdot BC \cdot r = \frac{1}{2} \cdot 10 \cdot \sqrt{93} = \boxed{5 \sqrt{93}}.]

All of you guys are wrong! The answer is actually D. 72. Bosco you're actually correct. The guest is just a troll and has no idea what he's saying. Therefore, the answer is still 72. I just found it out.

First, the word 'appreciated' means increased in value. If a house appreciates in value by 20%, it is now worth 20% more than its original price.

Let us call the original value of the rug \(x\). We know that the rug is worth 105% of what it was worth 10 years ago. We can write this as

\(12000 = 105\% \cdot x\). Converting the percent to a decimal gives us the equation \(12000=1.05x\). Dividing both sides by 1.05 gives us our final answer: \(x \approx \$11428.57\). (Rounded to the nearest cent.) So, the antique rug was worth approximately $11428.57.

This problem involves the addition of rational functions. Since fractions are involved, we have to consider when the denominator equals 0. This process is made easier by the process of factoring.

\(f(x) = \frac{x^2 + 10x + 21}{x^2 - 4x -21} + \frac{x^2 - 1}{x^2 - 4x + 4} \\ f(x) = \frac{x^2 + 10x + 21}{(x + 3)(x - 7)} + \frac{x^2 - 1}{(x - 2)^2}\)

After factoring the denominators, we can identify the x-values that will be outside of the domain.

\(x + 3 \neq 0 \quad x - 7 \neq 0 x - 2 \neq 0 \\ x \neq -3 \quad x\neq 7 \quad x \neq 2\)

All of these x-values are not within the domain.

Therefore, the domain in interval notation is \((-\infty, -3) \cup (-3, 2) \cup (2, 7) \cup (7, \infty)\)

\(f(x)\) is a rational function, so an x-value that could realistically not be in the domain occurs when the denominator equals 0.

\((x^2 - 6x + 8) - (x^2 + x - 6) = 0 \\ x^2 - 6x + 8 - x^2 - x + 6 = 0 \\ -7x+14 = 0 \\ -7x = -14 \\ x = -2\)

The value of x = -2 causes the denominator to be 0, x = -2 is the only x-value not in the domain of this function!

My method of solving this is to consider all the possible ways in which N could have a sum of digits no more than 4 and adding those together.

\(2 \times 10^7 \leq N < 10^8 \\ 20\;000\;000 \leq N < 100\;000\;000\)

Considering all the values of N is a tad overwhelming, so I started with limited my search to the range \(20\;000\;000 \leq N < 30\;000\;000\). How can the values of N within this range have a sum of digits no more than 4?

1) One way is to ensure all digits are 0. There is only 1 way this could happen.

2) One way is to ensure that one 1 appears in any digit except the leading digit. There are 7 digits available, so there are 7 ways this could happen.

3) One way is to ensure that 2 1's appear in any digit except the leading digit. There are 7 digits available for the first 1 and 6 remaining choices for the second 1. However, order is immaterial, so we must exclude the cases where different orders create the same number. \(\frac{7 * 6}{2} = 21 \text{ ways}\).

4) One way is to ensure that one 2 appears in any digit except the leading digit. There are 7 digits available, so there are 7 ways this could happen.

I then considered the range \(30\;000\;000 \leq N < 40\;000\;000\). Luckily, the cases are simpler than the previous range because the leading digit of 3 restricts some of the possibilities.

1) One way is to ensure all digits are 0. There is only 1 way this could happen.

2) One way is to ensure that one 1 appears in any digit except the leading digit. There are 7 digits available, so there are 7 ways this could happen. 

I then considered the range \(40\;000\;000 \leq N < 50\;000\;000\). This case is by far the simplest case.

1) One way is to ensure all digits are 0. There is only 1 way this could happen.

I then realized that there are no more options. For the range \(50\;000\;000 \leq N < 100\;000\;000\), there are no combinations of digits that meet the criteria since the leading digit exceeds the maximum sum.

Now, just add all the possible ways. \(1 + 7 + 21 + 7 + 1 + 7 + 1 = 45\) positive integers N that meet both criteria.

This problem just requires some clever algebraic manipulation. I know others have already posted the answer to this system of equations, but I will show you some work as to how one might find one of the variables because it may not be too straightforward of a process. If I can avoid it, I try to resist having fractions in a problem involving system of equations, so I decided to eliminate the fractions from the original equation. There is probably a faster way, but it is typically hard to find.

Afterwards, I solved for one of the variables in both equations. I chose to solve for y, but any choice should suffice. I used the technique of grouping all the terms with y and then factoring.

We have two equations solved for y, so we can find a relationship between x and z variables.

\(y = \frac{x}{x - 1}, y = \frac{4z}{z - 4} \\ \frac{x}{x - 1} = \frac{4z}{z - 4} \\ x(z - 4) = 4z(x - 1) \\ xz - 4x = 4xz - 4z \\ -3xz = 4x - 4z\)

Now, we can conveniently use the \(xz = 2x + 2z\) equation to help solve this system of equations.

\(xz = 2x + 2z \\ 3xz = 6x + 6z\)

Now, we can use technique of elimination to find a relationship between x and z.

\(\begin{align*} 3xz &= &6x &+ 6z \\ -3xz &= &4x &- 4z \\ 0 &= &10x &+ 2z \end{align*} \\ 5x + z = 0 \\ z = -5x\)

Now, we can substitute this relationship for z into an equation with x and z variables and solve for x.

\(xz = 2x + 2z \\ x * -5x = 2x + 2 * -5x \\ -5x^2 = -8x \\ -5x^2 + 8x = 0 \\ -x(5x - 8) = 0 \\ x = 0 \text{ or } 5x - 8 = 0 \\ x = 0 \text{ or } x = \frac{8}{5}\)

Now, you might notice that we obtained x = 0 as a potential solution, but we can reject \(x = 0\) immediately because if we were to substitute this x-value into the original equation \(\frac{xy}{x + y} = 1\), there would be no solution, so \(x = 0\) is not a candidate. At this point, I will not go any further because the process should be similar to find the other variables. Happy solving!

As Bosco discovered, this regular octagon cannot exist. It is a bit odd that a regular octagon problem would give the length of one of the sides and the perpendicular distance from one of the sides to the center of the octagon.

All the central angles in a regular polygon have a combined sum of \(360^{\circ}\) because they form a circle. In addition, all the central angles within a regular polygon have equal measure. Therefore, we can use division to find the number of sides, n.

\(n = \frac{360^{\circ}}{1^{\circ}} = 360\). This polygon has 360 sides!

This answer makes no sense. Your first sentence claims that Jessica sold 36 cookies and then you conclude somehow that Jessica sold 18 cookies. That is a major contradiction.

I am not familiar with the staar exam, but my review of Bosco's answer is that Bosco got the question right given his or her interpretation of this word problem. You all are claiming that A. 18 is the right answer without any explanation.