Question

math help

Answer 1

For question1, since the rope touches the top of the tree, then by similar triangles, the building must be
32 feet tall.

Answer 2

question 2 is B. BI = BK

This is true because perpendicular lines meet at a right angle.

Answer 3

i got those im at number 7

Answer 4

for question 3, the intersection of the perpendicular bisectors is (5, 5).

Therefore ,circumcenter is (5, 5) so it's A.

Answer 5

oh
wow i'm slow lol

Answer 6

so far you've been right though lol

Answer 7

Ok let me work on number 7 one second

Answer 8

okay

Answer 9

The only rule for if three lengths could be a triangle is the sum of the two smaller must be larger than the third, so.... it's B, because 12 + 9 = 21

Answer 10

for number 8, it has to be B

Answer 11

for number 9, it is A

Answer 12

number 10 it would be B since AB = BC

Answer 13

where are the letters for the fill in the blank? is there a word bank

Answer 14

no word bank

Answer 15

its just fill in

Answer 16

oh so you have to type in the word?

Answer 17

11. is midsegment

Answer 18

12. is equidistant

Answer 19

13. is perpendicular

Answer 20

14. is concurrent

Answer 21

3 more ?

Answer 22

For 15 i already found the answer lol:
B is the mid point of AC and D is the mid point of CE
In triangles CBD and CAE
CB/CA=1/2
CD/CE=1/2
angle C is the common angle
so these two triangles are similar
therefore all angles are equal , and all triangles are in the same ratio
hence BD/AE=1/2
BD=3x+5
AE=4x+20
3x+5/(4x+20)=1/2
6x+10=4x+20
2x=10
x=5

Just type that all in to show work! :)

Answer 23

For 16.

BE= BG+GE
BG is one/third of BE and that GE is two/thirds of BE
so that is 9 by 1/3 and 9 by 2/3
so the answer would be BG = 6 and GE = 3

Answer 24

oops my bad lol, BG = 3 and GE = 6

Answer 25

is there more questions?

Answer 26

yes just 3 more

Answer 27

ok post them

Answer 28

17. since presumably triangle DEF is similar to triangle DGF, and since side DF in triangle DEF is congruent to side DF in triangle DGF, then you can assume triangles DEF and DGF are congruent. So side FG is the same length as FE. You can set the expressions equal to each other where n + 9 = 4n - 6.
Solve for n:
9 = 4n - n - 6
9 = 3n - 6
9 + 6 = 3n
15 = 3n
15 / 3 = n
5 = n
Now substitute n for the expression that is the length of FG which is 4n - 6
4 * (5) - 6 = length FG
20 - 6 = length FG
14 = length FG

Answer 29

i did 20

Answer 30

type that all in to show your work for number 17

Answer 31

ok

Answer 32

for number 18, do you know how to sketch a graph of ABC?

Answer 33

wait, you don't have to show the work for sketching that right? just the steps to find the orthocenter?

Answer 34

yeah

Answer 35

we dont have to draw anything

Answer 36

Ok that's good Lol hold on let me post the steps for you

Answer 37

Since AB is a horizonal line, the altitude through C must be a vertical line. So, the orthocenter coordinates can be written as (1, b), and the direction of the altitude from A is (1, b-6), and the direction of BC is parallel to (1, 1). Since the dot product of two perpendicular vector is zero, we have

(1, b-6) ⋅ (1, 1) = 0

1 + b-6 = 0

b = 5

Answer: The orthocenter is at (1, 5).

Answer 38

For 19 just type these in :
-circumcenter
-incenter
-centroid
-orthocenter

Answer 39

last ones 21

Answer 40

ok 1 second

Answer 41

The incenter is equidistant from each side of the triangle.
The incenter is where all of the bisectors of the angles of the triangle meet.
The incenter of a triangle is always inside it.

Answer 42

that's it right?

Answer 43

yes thank you so much !
life saver

Answer 44

no problem, glad i could help!! i have to go now, bye :) good luck with the rest of your studies!! when's the last day to finish everything for the semester? my deadline is january 22

Answer 45

mines 19

Answer 46

but okay talk to you later

Answer 47

ohhh good luck, talk to you later :)