Each unit in the coordinate plane corresponds to 1 mile. Find the distance from the stadium to Pine Avenue. Round your answer to the nearest tenth.




The distance is about

miles.

Each Unit In The Coordinate Plane Corresponds To 1 Mile. Find The Distance From The Stadium To Pine Avenue.

Answers

Answer 1

The distance from the stadium to Pine Avenue is [tex]13.4miles[/tex], (to the nearest tenth)

The stadium is on Hazel Street(the green line). Pine Avenue is the blue line. The distance from the stadium to Pine Avenue will be the distance on the graph (in miles) from the point representing the stadium, to the point of intersection between Hazel Street and Pine Avenue.

We know the location of the stadium, [tex]S=(-6,6)[/tex]

The point of intersection can be obtained by solving the simultaneous equations

[tex]y=-\dfrac{1}{2}x+3\\y=2x-12[/tex]

since both equations are set equal to [tex]y[/tex], set them equal to each other, to eliminate [tex]y[/tex], and solve for [tex]x[/tex]

[tex]-\dfrac{1}{2}x+3=2x-12\\\\x=6[/tex]

Substitute this value of [tex]x[/tex], into the equation for Pine avenue

[tex]y-2x-12\\=2(6)-12\\=0[/tex]

The point of intersection between Hazel Street and Pine Avenue is [tex]I=(x,y)=(6,0)[/tex]

To compute the distance between the stadium and Pine Avenue, we find the distance between the two points [tex]S[/tex] and [tex]I[/tex]. So,

[tex]d=\sqrt{(6-(-6))^2+(0-6)^2}\\\\=\sqrt{12^2+(-6)^2}\\\\=6\sqrt5\approx13.4\text{ (to the nearest tenth)}[/tex]

The distance is [tex]13.4miles[/tex]

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Answers

Answer:

3) a = 9

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Step-by-step explanation:

Assuming here that the two questions you're talking about are problem 3 and problem 4, here is my work to find each answer.

Problem 3

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[tex]\huge\text{Hey there!}[/tex]

[tex]\huge\text{Number 1}[/tex]

[tex]\text{QUESTION: 5(a - 3) = 30}\\\\\large\textsf{DISTRIBUTE 5 WITHIN the PARENTHESES}\\\\\rm{5(a) + 5(-3) = 30}\\\\\rm{5a - 15 = 30}\\\\\large\textsf{ADD 15 to BOTH SIDES}\\\\\rm{5a - 15 + 15 = 30 + 15}\\\\\large\textsf{CANCEL out: -15 + 15 because it gives you 0}\\\large\textsf{KEEP: 30 + 15 because it helps solve for the a-value}\\\\\text{NEW EQUATION: 5a = 30 + 15}\\\\\large\textsf{SIMPLIFY IT!}\\\\\rm{5a = 45}\\\\\large\textsf{DIVIDE 5 to BOTH SIDES}\\\\\rm{\dfrac{5a}{5} = \dfrac{45}{5}}[/tex]

[tex]\large\textsf{CANCEL out: }\mathsf{\dfrac{5}{5}}\large\textsf{ because it gives you 1}\\\large\textsf{KEEP: }\mathsf{\dfrac{45}{5}}\large\textsf{ because it helps solve for the a-value}\\\\\text{YOUR NEW EQUATION: }\rm{a = \dfrac{45}{5}}\\\\\large\textsf{SIMPLIFY IT!}\\\\\large\text{a = 9}\\\\\huge\text{Therefore, your answer is: \boxed{\textsf{a = 9}}}\huge\checkmark[/tex]

[tex]\huge\text{NUMBER 2}[/tex]

[tex]\text{QUESTION: }\rm{\dfrac{v}{2} + 2 = 10}\\\\\rm{\rightarrow \dfrac{1}{2}v + 2 = 10}\\\\\large\textsf{SUBTRACT 2 to BOTH SIDES}\\\\\rm{\dfrac{1}{2}v + 2 - 2 = 10 - 2}\\\\\large\textsf{CANCEL out: 2 - 2 because it gives you 0}\\\large\textsf{KEEP: 10 - 2 because it helps solve for the v-value.}\\\\\text{YOUR NEW EQUATION: }\rm{\dfrac{1}{2}v = 10 - 2}\\\\\text{SIMPLIFY IT!}\\\\\rm{\dfrac{1}{2}v = 8}\\\\\large\textsf{MULTIPLY 2 to BOTH SIDES}\\\\\rm{2\times \dfrac{1}{2}v = 2\times8}[/tex]

[tex]\large\textsf{CANCEL out: }\mathsf{2\times\dfrac{1}{2}}\large\textsf{ because it gives you 1}\\\large\textsf{KEEP: }\mathsf{2\times8}\large\textsf{ because it helps solve for the v-value}\\\\\text{YOUR NEW EQUATION: }\rm{v = 2\times8}\\\\\large\textsf{SIMPLIFY IT!}\\\\\rm{v = 16}\\\\\huge\text{Therefore, your answer is: \boxed{\textsf{v = 16}}}\huge\checkmark[/tex]

[tex]\huge\text{Good luck on your assignment \& enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

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Answer:

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Answers

The solution to the system of equations is (2,0)

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