Question

**#4.) 1.06: in the diagram, wx is the segment bisector of yz. yt = 2x + 10, wx = 6x - 1, and yz = 60. determine the length of zt.
**#5.) 1.05: solve for dg in the diagram below.
d diagram with points e, f on line dg, lengths 12, 11x, 14x + 3, 21 shown
*#6.) 1.06: points m, n, and o are on a line where m is the mid - point. if no = 2x² - 16 units and mn = 2x, determine the length of mo.
a. 8 units
b. 4 units
c. 32 units
d. 16 units
#7.) unit 1 is all about learning the foundations of geometry that we will be using all year long. with your mid - unit quiz coming up, which topic(s) do you feel most comfortable with? which do you need more help with? why?

Explanation:

Step1: Use segment - bisector property for #4

Since $WX$ is the segment bisector of $YZ$, then $YT = TZ$ and $YZ=YT + TZ$. Given $YZ = 60$, so $YT=\frac{YZ}{2}=30$. Also, $YT = 2x+10$. Set up the equation $2x + 10=30$.
\[2x+10 = 30\]
\[2x=30 - 10\]
\[2x=20\]
\[x = 10\]
Since $YT=TZ$ and $YT = 30$, then $ZT = 30$.

Step2: Use segment - length relationships for #5

From the diagram, $DG=14x + 3$ and also $DG=11x+12$. Set up the equation $14x + 3=11x+12$.
\[14x-11x=12 - 3\]
\[3x=9\]
\[x = 3\]
Substitute $x = 3$ into the expression for $DG$, $DG=14x+3=14\times3+3=42 + 3=45$.

Step3: Use mid - point property for #6

Since $M$ is the mid - point of $\overline{NO}$, then $MN=MO$. Given $NO = 2x^{2}-16$ and $MN = 2x$, and $NO=MN + MO=2MN$ (because $MN = MO$). So $2x^{2}-16=2\times(2x)$.
\[2x^{2}-16 = 4x\]
\[x^{2}-2x - 8=0\]
Factor the quadratic equation: $(x - 4)(x+2)=0$. So $x=4$ or $x=-2$. Since length cannot be negative, we take $x = 4$. Then $MO=MN=2x=8$.

Answer:

#4: $30$
#5: $45$
#6: A. 8 units