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$Math question 3 a) workout: {:[" Brisbane "=(b_(1),b_(2))],[13*34=sqrt((b_(1)-22)^(2)+(b_(2)-2)^(2))],[(13.34)^(2)=(sqrt((b)-22)^(2)+(b_(2)-2))^(2)],[178=(b_(1)-22)^(2)+(b_(2)-2)^(2)],[a^(2)+b^(2)=178=b_(1)-22)^(2)+(b_(2)-2:}],[3^(2)+13^(2)=178=(b_(1)-22)^(2)+(b_(2):}],[3=b-22],[+22],[3+22=b quad∣b=25]:} {:[13=b_(2)-2],[+2+2],[13+2=b_(2)quad(b_(2)=15:}],[" Brisbane "=(25","15)]:}$

Math question $3$ a) workout: $\newline$ $\begin{array}{l} \text { Brisbane }=\left(b_{1}, b_{2}\right) \\ 13 \cdot 34=\sqrt{\left(b_{1}-22\right)^{2}+\left(b_{2}-2\right)^{2}} \\ \left.(13.34)^{2}=(\sqrt{(b}-22)^{2}+\left(b_{2}-2\right)\right)^{2} \\ 178=\left(b_{1}-22\right)^{2}+\left(b_{2}-2\right)^{2} \\ \left.a^{2}+b^{2}=178=b_{1}-22\right)^{2}+\left(b_{2}-2\right. \\ 3^{2}+13^{2}=178=\left(b_{1}-22\right)^{2}+\left(b_{2}\right. \\ 3=b-22 \\ +22 \\ 3+22=b \quad \mid b=25 \end{array}$ $\newline$ $\begin{array}{l} 13=b_{2}-2 \\ +2+2 \\ 13+2=b_{2} \quad\left(b_{2}=15\right. \\ \text { Brisbane }=(25,15) \end{array}$

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Grade 8

Identify equivalent linear expressions: word problems

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Q. Math question

3

a) workout:

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\begin{array}{l} \text { Brisbane }=\left(b_{1}, b_{2}\right) \\ 13 \cdot 34=\sqrt{\left(b_{1}-22\right)^{2}+\left(b_{2}-2\right)^{2}} \\ \left.(13.34)^{2}=(\sqrt{(b}-22)^{2}+\left(b_{2}-2\right)\right)^{2} \\ 178=\left(b_{1}-22\right)^{2}+\left(b_{2}-2\right)^{2} \\ \left.a^{2}+b^{2}=178=b_{1}-22\right)^{2}+\left(b_{2}-2\right. \\ 3^{2}+13^{2}=178=\left(b_{1}-22\right)^{2}+\left(b_{2}\right. \\ 3=b-22 \\ +22 \\ 3+22=b \quad \mid b=25 \end{array}

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\begin{array}{l} 13=b_{2}-2 \\ +2+2 \\ 13+2=b_{2} \quad\left(b_{2}=15\right. \\ \text { Brisbane }=(25,15) \end{array}

Square Distance Equation: First, let's square the distance equation to get rid of the square root. $\newline$ $(13 \times 34)^2 = ((b_1 - 22)^2 + (b_2 - 2)^2)$
Calculate Left Side: Now, calculate the left side of the equation. $\newline$ $(13\times34)^2 = 178^2$
Write Right Side: Next, write down the right side of the equation. $\newline$ $178^2 = (b_1 - 22)^2 + (b_2 - 2)^2$
Use Pythagorean Theorem: We have another equation given by $3^2 + 13^2 = 178 = (b_1 - 22)^2 + (b_2 - 2)^2$ .
Solve for $b_1$ : Now, let's solve for $b_1$ using the equation $3 = b_1 - 22$ . $\newline$ $3 = b_1 - 22$
Find $b_1$ : Add $22$ to both sides to find $b_1$ . $\newline$ $3 + 22 = b_1$ $\newline$ $b_1 = 25$
Solve for $b_2$ : Now, let's solve for $b_2$ using the equation $13 = b_2 - 2$ . $\newline$ $13 = b_2 - 2$
Find $b_2$ : Add $2$ to both sides to find $b_2$ . $\newline$ $13 + 2 = b_2$ $\newline$ $b_2 = 15$
Final Coordinates: Finally, write down the coordinates of Brisbane. ＂Brisbane＂ = $(25, 15)$