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Hunter-Best [27]
3 years ago
12

On a map of a town, a school is located at (–6, 2) and a movie theater is located at (–6, –15). What is the distance from the sc

hool to the theater on the map?

Mathematics
2 answers:
Anit [1.1K]3 years ago
5 0

Answer:

Step-by-step explanation:

vesna_86 [32]3 years ago
3 0

Answer:

13 units

Step-by-step explanation:

Since the x-coordinates are the same, you can simply add the absolute value of the y-coordinates.

|2 + (-15)|

or

|2 - 15|

= 13

Therefore, the distance from the school to theater is 14 units.

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What is the value of a3 in the sequence?
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Basically, we are trying to find the third number in the pattern. The pattern is - -4, -2, -12, -14 and so on. We can see that the third number is -12, which is a₃. Therefore, the answer is C) -12. Hope this helped!
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A store owner will order iPads for local schools. The relationship between x, the number of boxes he will order, and y, the numb
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Answer:

y, ‎y = ‎15x.

Step-by-step explanation:

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2 years ago
Find the sum of the three expressions and choose the correct answer.
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(-2u^3v + 5uv - 1) +(7u^3v - 2u^2v^2) +(5u^2v^2 - uv + 6)=\\\\ -2u^3v + 5uv - 1 +7u^3v - 2u^2v^2 +5u^2v^2 - uv + 6= \\ \\ (7u^3v-2u^3v)+(5u^2v^2-2u^2v^2)+(5uv-uv)+(6-1)=\\\\ \boxed{5u^3v+3u^2v^2+4uv+5}
5 0
3 years ago
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PLEASE HELP ME:<br><br> Find the perimeter of the figure. Round to the nearest tenth.
Ksenya-84 [330]

Answer:

perimeter = 20.9 units

Step-by-step explanation:

perimeter

perimeter = distance around two dimensional shape

= addition of all sides lengths

<h2>perimeter of the figure</h2><h2>= AB+BC+CD+AD</h2>

distance formula:

d =  \sqrt{(  x_{2} -  x_{1})  {}^{2}   + ( y_{2} -  y_{1}) {}^{2}  }

<h3>1) distance of AB</h3>

A(-3,0) B(2,4)

x1 = -3 x2 = 2

y1 = 0 y2 = 4

(substitute the values into the distance formula)

ab =  \sqrt{(2 - ( - 3)) {}^{2}  + (4 - 0) {}^{2} }

ab =  \sqrt{5 {}^{2} + 4 {}^{2}  }

ab =  \sqrt{41}AB = 6.4 units

<h3>2) distance of BC</h3>

B(2,4) C(3,1)

x1 = 2 x2 = 3

y1 = 4 y2 = 1

bc =  \sqrt{(3 - 2) {}^{2}  + (1 - 4) {}^{2} }

bc = \sqrt{1 {}^{2} + ( - 3) {}^{2}  }

bc =  \sqrt{10}

BC = 3.2 units

<h3>3) distance of CD</h3>

C(3,1) D(-4,-3)

x1 = 3 x2 = -4

y1 = 1 y2 = -3

cd =  \sqrt{( - 4 - 3) {}^{2}  + ( - 3 - 1)) {}^{2} }

cd =  \sqrt{( -   7) {}^{2}  + ( - 4 ){}^{2} }

cd =   \sqrt{65}

CD = 8.1 units

<h3>4) distance of AD</h3>

A(-3,0) D(-4,-3)

x1 = -3 x2 = -4

y1 = 0 y2 = -3

ad =  \sqrt{(  - 4 - ( - 3)) {}^{2}  +  ( - 3 - 0) {}^{2} }

ad =  \sqrt{( - 1) {}^{2} + ( - 3) {}^{2}  }

ad =  \sqrt{10}

AD = 3.2 units

<h2>perimeter of figure</h2>

= AB+BC+CD+AD

= 6.4 + 3.2 + 8.1 + 3.2

= 20.9 units

8 0
3 years ago
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