Question

Your itinerary contains a grid map of Sydney, with each unit on the grid representing 0.125 kilometers. If the Sydney Opera House is located at (−9,−9) and the Queen Victoria Building is located at (−2,10), what is the direct distance (not walking distance, which would have to account for bridges and roadways) between the two landmarks in kilometers? Round your answer to two decimal places, if necessary.

Answer 1

distance formula between the two points

small catch: because each unit on the grid represents 0.125 km, you also need to multiply your result by 0.125

Answer 2

(-9, -9), (-2, 10) ----> (-2, -9), (10, -9)

Answer 3

i'm trying to figure it out how to put it in the formula √

Answer 4

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

let (x1,y1) be (-9, -9)
let (x2,y1) be (-2, 10)

\[d = \sqrt{(-2-(-9))^{2}+((10-(-9))^{2}}\]

Answer 5

\[\sqrt{410}\]

Answer 6

good, sqrt(410) units

since 1 unit = 0.125km, we multiply sqrt(410) * 0.125 to get 2.53km

Answer 7

Given (16,11) and (x,−4), find all x such that the distance between these two points is 17. Separate multiple answers with a comma.

x =

Answer 8

Given (16,11) and (x,−4), find all x such that the distance between these two points is 17. Separate multiple answers with a comma.

x =

Answer 9

distance formula again

plug in the points. however, this time you set the distance equal to 17 and solve for x. there should be 2 solutions.

Answer 10

\[\sqrt{x^2 - 32x + 481}\]

Answer 11

oh wait

Answer 12

good so far

set that equal to 17 and solve

square both sides, subtract 17^2 from both sides, factor or use the quadratic formula

Answer 13

\[\sqrt{x^2 - 32x + 481} - 289\]

Answer 14

when you square both sides, the radical sign disappears

17 = sqrt(x^2 - 32x + 481)

17^2 = x^2 - 32x + 481

subtracting 17^2 from both sides

x^2 - 32x + 192 = 0

factoring: (x-8)(x-24) = 0

therefore x = 8 and x = 24

Answer 15

oh okay

Answer 16

Consider the following points.

(−3,8),(−8,2) and (2,2)

Step 1 of 2 : Determine whether or not the given points form a right triangle. If the triangle is not a right triangle, determine if it is isosceles or scalene.

Answer 17

@surjithayer would you mind taking a look at this? getting a little tired

Answer 18

i do you one better

Answer 19

it is a isosceles triangle ?

Answer 20

Thank you vocaloid for all your help today , i truly needed it and appreciate it

Answer 21

horizontal distance=-2-(-9)=7 units=7*0.125=0.875 km
vertical distance=10-9=1 unit=0.125 km
direct distance\[=\sqrt{0.875^2+(0.125)^2}\approx0.88388 ~km \approx 0.88 km\]

Answer 22

0.125 is not diagonal or slant it is vertical or horizontal distance.
what i think,i may be wrong.

Answer 23

\[17=\sqrt{(x-16)^2+(-4-11)^2}\]
\[(x-16)^2+225=289\]
\[(x-16)^2=289-225\]
\[(x-16)^2=49\]
\[x-16=\pm7\]
x-16=7
x=7+16=23
or
x-16=-7
x=-7+16=9
x=9,23

Answer 24

let the points be A(-3,8),B(-8,2),C(2,2)
slope of AB\[=\frac{ 2-8 }{ -8+3}=\frac{ -6 }{- 5 }=\frac{6}{5}\]
slope of BC\[=\frac{ 2-2 }{ 2+8 }=0\]
\[BC \left| \right|x-axis\]
slope of AC\[=\frac{ 2-8 }{ 2+3 }=-\frac{ 6 }{ 5 }\]

Answer 25

as product of two slopes\[\neq-1\]
hence it is not a right angled triangle.

Answer 26

\[AB=\sqrt{(-8+3)^2+(2-8)^2}=\sqrt{25+36}=\sqrt{61}\]
\[BC=\sqrt{(2-2)^2+(2+8)^2}=\sqrt{0+100}=10\]
\[AC=\sqrt{(2+3)^2+(2-8)^2}=\sqrt{25+36}=\sqrt{61}\]
AB=AC
Hence it is an isosceles triangle.

Answer 27

that's right , it was

Answer 28

Thank you surj for your excellence and explanations

Answer 29

yw