Question

Hey! I did a bit but I need a couple more problems :/ I don't understand this one at all..

Answer 1

@jigglypuff314

Answer 2

http://www.mathopenref.com/constparallel.html

Answer 3

That's not the same thing. @jigglypuff314

Answer 4

Hello, Nick,

The illustration shows the y-axis, the point P to the right of the y-axis, and a line drawn through P.

Here are a couple of ways of responding to this question:

(1) Using P as your axis, rotate the line through P until it is exactly parallel to the y-axis. If you really wanted to be fussy, you could take a good ruler and use it to ensure that every point on the line thru P is the same distance from the y-axis.

(2) You could measure how far P is from the y-axis in the horizontal direction (perpendicular to the y axis or parallel to the x-axis). Then mark another point Q, either farther from the x-axis or closer to it than P), and then reposition the existing line through P so that it goes through the new point Q as well.

Answer 5

So the answer would be "Measure the distance from the intersection of the transversal to Point P".

Answer 6

A lot depends on what your teacher's expectations are. Were I your teacher, I'd accept a verbal answer such as one of those I've typed above.

Answer 7

@jigglypuff314 :) You're such a big help!

Answer 8

Nick: if you're allowed (or encouraged) to use mathematical methods to answer questions such as these, I'd suggest you apply the Pythagorean Theorem. You have a triangle with the lengths of the legs known and are to find the length of the hypotenuse.

Answer 9

\[distance formula: \sqrt{(x _{2}-x _{1})^{2} + (y _{2}-y _{1})^{2}}\]given points (x1, y1) and (x2, y2)

Answer 10

JigglyPuff: You really are a big help. Your formula is exactly right for this situation.

Answer 11

thanks :)
it's better to go short and sweet, kids don't like reading paragraphs these days ;)

Answer 12

5.83 :) ew. paragraphs *shivers*

Answer 13

lol and correct :)

Answer 14

Sorry about being AFK im just trying to solve a bit on my own. I don't want to just ask for answers all the time :)

Answer 15

don't worry ;) you being able to do a lot of it by yourself is a good thing :)

Answer 16

A triangle can be formed with side lengths 120,85,and 220 True or false?

Answer 17

120 + 85 = 205
120 - 85 = 35
220 is Not between 35 and 205

Answer 18

Got it ;) Classify the triangle based on it's side lengths 5,12, and 13.

Answer 19

you can try...
does 5^2 + 12^2 = 13^2?
if it works, then it's a right triangle...

Answer 20

Got it! :)

Answer 21

area of trapezoid: (1/2)h(b1+b2)
in other words, average of bases times height.

Answer 22

What...?

Answer 23

(11.75 + 18)(1/2)(6.12) =

Answer 24

91.035 :) This question is different.

Answer 25

Would it be 119?

Answer 26

not quite...
add the area of the trapezoid to the area of the triangle
the area of the triangle would be = (1/2)(12)(13.42) =

Answer 27

80.52?

Answer 28

then add that to the answer you got before

Answer 29

171.555

Answer 30

That's what I got :)

Answer 31

:)

Answer 32

30x + 11y = 112
2x + 3y = 12

Answer 33

I get y=0 but that is wrong :/

Answer 34

I solved this by the elimination method
30x + 11y = 112
2x + 3y = 12 (multiply through by 15)
-> 30x + 45y = 180
- (30x + 11y = 112)
-> 0 + 34y = 68 then divide both sides by 34
then plug what you get for y back into 2x + 3y = 12

Answer 35

y=2?

Answer 36

and x=3 :)

Answer 37

yep :)

Answer 38

:D:D:D:DD

Answer 39

the diagonals bisect each other
so to get one side it would be... √(1.5^2 + 2^2)

Answer 40

10? :)))))))))))))

Answer 41

correct :)
and
DG = AB
angle GDB = angle BAD
DB = DB

Answer 42

SAS

Answer 43

yep :)

Answer 44

Hi, Nick! Glad you're making progress. How about posting no more than 1 or 2 problems in any one separate problem solving session?

Answer 45

Got it jiggly

Answer 46

so we all get a chance of earning more medals! :D
lol jk
It's one of the site rules...
I think it so that moderators can check up on things better
and so in case reinforcements have to be called to help on a problem they wouldn't have to read through everything 0.o I had to do that once... not fun...

Answer 47

Should we go to another one ? :/ or just stay?

Answer 48

how about make it a habit to just post each question separately ;)

Answer 49

Ok ;3

Answer 50

height = √(20^2 - 12^2) =
area = base times height

Answer 51

Wrong one :3

Answer 52

I forget what it's called >,<
like the one like if the triangles are congruent then its angles are congruent...

Answer 53

SSA :3

Answer 54

whoops looks like I was thinking too far ahead :P
not SSA though... SSA is not a true postulate.
try
QU = AD
QD = AU
DU = DU

Answer 55

SSS

Answer 56

yep :)

Answer 57

AAS

Answer 58

angle TAM = angle MAH
MA = MA
angle TMA = angle HMA

Answer 59

well I just got confuzeled... >,<
idk this one (got cross-eyed)

Answer 60

A. Not Congruent
B.SSA
C.HL
D.SAS

Answer 61

A!!!

Answer 62

Wait really

Answer 63

\[\Huge\bf\color{#FF0000}{A}\]yeah it was why I was confused,
cuz I was like... "but... but... they're not congruent D,x"

Answer 64

hahaha xD

Answer 65

area = length times width
perimeter = 2L + 2W
38 = 2(10) + 2W solve for W and use it in area formula

Answer 66

90 :D

Answer 67

yep :)

Answer 68

I'll do a bit on my own :) after this one :3

Answer 69

c = √ (5√2)^2 + (5√2)^2 )

Answer 70

Idk what that is :O

Answer 71

(5√2)^2 = 50
50 + 50 = 100
c = √100 =

Answer 72

10.. I still don't understand that one :/

Answer 73

it was the triangle rule things...
so for the question... x = 14/2 √3