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