Question

coordinate help again

Answer 1

the first is the same as last one, except you solve for a side instead of the longer slant

side^2 + side^2 = slant^2

s^2 + 1^2 = root(2)^2

s^2 + 1 = 2

Answer 2

this is a special one, if the sides are the same , and the slant is root 2 times the side length, it is a 45-45-90 triangle

Answer 3

hmm ok?

Answer 4

im lost

Answer 5

solve s^2+1^2=root(2)^2

s^2 + 1 = 2

s^2 = 1

s=1

Answer 6

the sides are in a ratio of 1 to 1 to root (2), it is a 45-45-90 triangle, as an aside

Answer 7

remember that for a right triangle..

a^2 + b^2 = c^2

a and b are the sides and c is the hypotenuse

Answer 8

so 1 is th emissing length

Answer 9

s^2+1=2?

Answer 10

yeah s is a variable i chose for the missing side

Answer 11

okay okay by for c=2 its sqrt right?

Answer 12

\[c=\sqrt{2}\]
\[c^2=[\sqrt{2}]^2=2\]

Answer 13

?

Answer 14

pluggin in you get

a^2 + b^2 = c^2

a^2 + 1 = 2

Answer 15

a=1?

Answer 16

The thing to remember is for any right triangle... the sum of the squares of the two sides, 'a' and 'b' is the same as the square of the hypotenuse

a^2 + b^2 = c^2

Answer 17

here they give you a side length, and the hypotenuse length , putting those in you get

a^2 + 1^2 = 2

a^2 = 1

a=1

Answer 18

so A is the correct answer choice?

Answer 19

yes

Answer 20

you just have to know the a^2+b^2=c^2, the rest is just putting in the values ad solving for the missing one, here it is 1

Answer 21

okay thanks and what about the other question

Answer 22

THe figure in the second has pairs of vertical angles where the diagonals cross

recall vertical angles are the same measure, (opposite sides of an x are vertical angles)


so you can set those equal

Answer 23

yeah

Answer 24

4x - 2 = x + 28
and
4y - 7 = y + 14

Answer 25

that's it, just solve for the x and y values

Answer 26

10 and 7?

Answer 27

yep

Answer 28

thank you so much