select the correct answer.\nterbium - 147 undergoes positron emission to become a stable atom. what is that…

select the correct answer.\nterbium - 147 undergoes positron emission to become a stable atom. what is that stable atom?\n$_{65}^{147}tb\\to\\_\\_\\_+_{1}^{0}e$\na. $_{65}^{147}tb$\nb. $_{67}^{151}ho$\nc. $_{69}^{155}tm$\nd. $_{64}^{147}gd$\ne. $_{68}^{151}er$\nf. $_{70}^{151}yb$

select the correct answer.\nterbium - 147 undergoes positron emission to become a stable atom. what is that stable atom?\n$_{65}^{147}tb\\to\\_\\_\\_+_{1}^{0}e$\na. $_{65}^{147}tb$\nb. $_{67}^{151}ho$\nc. $_{69}^{155}tm$\nd. $_{64}^{147}gd$\ne. $_{68}^{151}er$\nf. $_{70}^{151}yb$

Answer

Explanation:

Step1: Recall slope formula

The slope formula between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $m=\frac{y_2 - y_1}{x_2 - x_1}$.

Step2: Calculate slope of Line 1

For Line 1 with points $(1,-3)$ and $(10,-3)$, substitute $x_1 = 1,y_1=-3,x_2 = 10,y_2=-3$ into the slope formula: $m_1=\frac{-3-(-3)}{10 - 1}=\frac{-3 + 3}{9}=0$

Step3: Calculate slope of Line 2

For Line 2 with points $(6,-9)$ and $(-9,-9)$, substitute $x_1 = 6,y_1=-9,x_2=-9,y_2=-9$ into the slope formula: $m_2=\frac{-9-(-9)}{-9 - 6}=\frac{-9 + 9}{-15}=0$

Step4: Determine relationship between lines

Since $m_1 = m_2=0$, the two lines are parallel.

Answer:

The slope of Line 1 is $m = 0$ The slope of Line 2 is $m = 0$ Line 1 is parallel to Line 2