dierdurst

I think this is a past asked question. Maybe it would be helpful to check it out here!

It is possible that they understood how to solve it and finished the problem for you.

- Daisy

6. \(3| x - 3 | < 12\)

First, divide both sides by \(3\) to get \(| x - 3 | < 4\). Now we have that \(x - 3 < 4\) or that \( x - 3 > -4\). By solving each one, we get \(x<7\) or \(x>-1\). This gives us \((-1, 7)\).

7. \( 5| 2x + 8 | + 4 < 9\)

First, subtract \(4\) from both sides to get \(5\left|2x+8\right|<5\). Then, we can divide by \(5\) on both sides to get \(\left|2x+8\right|<1\). Now, we have \(-1<2x+8<1\). By subtracting \(8\) on both sides then dividing by \(2\), we get \(x>-\frac{9}{2}\) and \(x<-\frac{7}{2}\) so we have \((-\frac{9}{2},-\frac{7}{2})\).

8. \(-2| 8x - 4 | + 1 < -15\)

Subtract \(1\) from both sides to get \(-2\left|8x-4\right|<-16\). Then, divide by \(-2\) on both sides to get \(\left|8x-4\right|>8\). Now, we can get that \(8x-4>8\) and \(8x-4<-8\). By simplifying both inequalities, we get \(\:\left(-\infty \:,\:-\frac{1}{2}\right)\cup \left(\frac{3}{2},\:\infty \:\right)\). 

- Daisy

To reduce the fraction, we need to find how many \(115\)'s are in \(247\). There are \(2\). \(115*2=230\). So \(247-230=17\) which is the remainder. Therefore, we have \(-2 \frac{17}{115}\). (Don't forget the negative!)

- Daisy

By using Complementary Counting, you can find the probability that no days snow, and subtract that from \(1\). The probability that no day snows is \(\frac{1}{3}*\frac{1}{3}*\frac{1}{3}=\frac{1}{27}\), so the probability of it snowing at least one day is \(1 - \frac{1}{27}= \frac{26}{27}\).

- Daisy

The diagonal of the rectangle would be the radius, so you can use Pythagorean Theorum to find the diagonal. We have \(6^2+8^2=(2x)^2\), so \(100=4r^2\), \(r=5\).

- Daisy

If the first foci is \((3,7) \) and the second is \((d, 7)\), then \(C = \) \((\frac{d+3}{2}, 7)\). \(C\) is the center of the ellipse. The point where the ellipse touches the x-axis is \((\frac{d+3}{2}, 0)\).

We know that \(P\)(any points on the ellipse) = \(T\), so

\(2 \sqrt{\left( \frac{d - 3}{2} \right)^2 + 7^2} = d + 3\)  

\(\sqrt{\left( \frac{d - 3}{2} \right)^2(4) + 7^2(4)} = d + 3\)

\(\sqrt{(d - 3)^2 + 196} = d + 3\)

Then, by squaring both sides, we get 

\((d - 3)^2 + 196 = d^2 + 6d + 9\)

\(d^2-6d+9+196=d^2+6d+9\)

\(12d=196\)

\(d=\frac{49}{3}\)

- Daisy

No problem! 

-Daisy

Hi CPhill, I think you can factor the last trinomial into \((x+2y)^2\) and get \((x+2y)^3 \)as a final answer. Haha, but great job! Thanks for explaining it!

-Daisy

Thank you! I think you made a slight mistake because we have (-2,3) instead of (2,3). Regardless, thank you for your help!

-Daisy

1a. \(6x^2+5x-4\)

We can split the middle by having \(6*(-4)=(-24)\). Now, we need to find factors of \(-24\) that add to \(5\). This is \(8\) and \(-3\). So we have \(6x^2-3x+8x-4\). Note that this does not change the value of the equation. Now, we can factor to get \(3x(2x-1)+4(2x-1)\) so we get \((3x+4)(2x-1)\).

1b. \(5x^2-4x-1\). 

\(5*(-1)=(-5)\). Factors of \(-5\) include \(-5\) and \(1\). We now can rewrite the equation as \(5x^2-5x+x-1\), which factors into \(5x(x-1)+(x-1)\), so we can get \((x-1)(5x+1)\).

2a. \(x-x^3\)

We can factor out \(x\) first, which gives us \(x(1-x^2)\). Then we can use difference of squares which gives us \(x(1-x)(1+x)\).

2b. \(x^3+2x^2+x\)

We can first factor out \(x\), which gives us \(x(x^2+2x+1)\). Then, we can factor the inside trinomial into \(x(x+1)^2\)

2c. \(x^6+4x^3-5\)

We can split the middle to get \( x^6-3x^3-x^3-5,\) which we can factor into \((x^2+5)(x^3-1)\). Then, we can use difference of cubes on the second term to get \((x^3+5)(x-1)(x^2+x+1)\)

2d. \(16x^4-1\)

We can rewrite this as \((4x)^2-(1)^2\), which we can use difference of squares on to get \((4x^2-1)(4x^2+1)\). Then, we can use difference of squares again to get \((2x-1)(2x+1)(4x^2+1)\)

3a. \(2xy+x+2y+1\)

We can factor out \(x\) from \(2xy+x\). This gives us \(x(2y+1)+(2y+1)\). Then, we get \((x+1)(2y+1)\)

3b. \(x^3+ax^2-x-a\)

We can factor out \(x^2\) from the first two terms. This gives us \(x^2(x+a)-(x+a)\). Then, we get \((x+a)(x^2-1) \)which we can use difference of squares to get \((x+a)(x-1)(x+1)\)

3c. \(x^3+6x^2y+12xy^2+8y^3\)

I'm actually not quite sure about this one, if someone else can do it that would be great. Sorry! 

3d. \(x^2+y^2+z^2+2xy-2xz-2yz\)

We can rearrange this and factor some of it to get \((x+y)^2-2z(x+y)+z^2\). Then, we can see that this is a perfect square trinomial and we can factor it to get \((x+y-z)^2\)

-Daisy

Let's use a Stars and Bars approach on this. There are 30 votes(stars) and 4 choices, but to get 4 catergories we only need 3 bars. In order to find where these bars would go, we can do 33C3 which is 5456.

- Daisy

You can draw triangle \(PAS\) with \(DS\) perpendicular(the altitude) to \(AP\). You know \(PS = 10\). Let's name the midpoint of \(PS\) the letter \(F\). \(PF = 5, PA = 13, \) so \(AF = 12\). Now, cos \(PAS = \frac{AD}{AS}\). We already know \(AS = 13,\) so we just need to solve for \(AD\). To find \(AD\), we first need to solve for \(DS\), which is one of the altitudes of triangle \(PAS\). We know that the area of a triangle is \(\frac{1}{2}bh\), so we can make an equation. \(\frac{1}{2}\times10\times12=\frac{1}{2}\times13\times DS\). From this, we get \(DS = \frac{120}{13}\). Now, we can use Pythagorean Theorum to find \(AD\). \(12^2-(\frac{120}{13})^2=AD^2\). After calculating this, we can get \(AD = \frac{12\sqrt{69}}{13}\). Now, lets plug this into our fraction. We can get \(\frac{\frac{12\sqrt{69}}{13}}{13}\) which simplifies to \(\frac{12\sqrt{69}}{169}\).

- Daisy