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beks73 [17]
1 year ago
4

I will give brainliest to anyone who answer this right with a clear solution. Pls help ASAP. Thank you in advance.​

Mathematics
2 answers:
nordsb [41]1 year ago
6 0

Answer:

31.  Omission of the square root sign in step 1.

32.  Subtracting the y-value from the x-value in each parentheses in step 1.

Step-by-step explanation:

\boxed{\begin{minipage}{7.4 cm}\underline{Distance between two points}\\\\$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$\\\\\\where $(x_1,y_1)$ and $(x_2,y_2)$ are the two points.\\\end{minipage}}

Given points:

  • A = (6, 2)
  • B = (1, -4)

<h3><u>Question 31</u></h3>

The error was the omission of the square root sign in the first step of the calculation:

\textsf{Error}: \quad AB=(6-2)^2+(1-(-4))^2

\textsf{Correction}: \quad AB=\sqrt{(6-2)^2+(1-(-4))^2}

<u>Correct calculation</u>:

\begin{aligned}AB&=\sqrt{(6-1)^2+(2-(-4))^2}\\&=\sqrt{5^2+6^2}\\&=\sqrt{25+36}\\&=\sqrt{61}\\& \approx 7.8\end{aligned}

<h3><u>Question 32</u></h3>

The error was subtracting the y-value from the x-value in each parentheses in the first step of the calculation, rather than subtracting the x-values and the y-values separately:

\begin{aligned}\textsf{Error}: \quad AB&=\sqrt{(x_A-y_A)^2+(x_B-y_B)^2}\\ &=\sqrt{(6-2)^2+(1-(-4))^2}\end{aligned}

\begin{aligned}\textsf{Correction}: \quad AB&=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}\\&=\sqrt{(6-1)^2+(2-(-4))^2\end{aligned}

<u>Correct calculation</u>:

\begin{aligned}AB&=\sqrt{(6-1)^2+(2-(-4))^2}\\&=\sqrt{5^2+6^2}\\&=\sqrt{25+36}\\&=\sqrt{61}\\& \approx 7.8\end{aligned}

DanielleElmas [232]1 year ago
6 0

Answer:

7.81

Step-by-step explanation:

x1= 1

x2= 6

y1= -4

y2= 2

31.The first one just has one minor error which is not putting a square root.

after putting the square root on  \sqrt{61} you will get an answer which is 7.81 units

the second one has an error in putting the correct values

32. the equation for distance is \sqrt{(x_2-x_1 )+(y_2-y_1)} } however the values are inputted wrong as x2=6 and x1=1 but here x1 is taken as 2 which is the value of y2 and instead of inputting y2 as 2 it is written as 1 which is the value of x1

<u><em>Comment if you still don't understand</em></u>

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