Question

more math help ss below

Answer 1

@imagine

Answer 2

@extrinix

Answer 3

@snowflake0531

Answer 4

@iimaddii

Answer 5

@mrmudd183 can you help me

Answer 6

this is all about circle theorems -- the one we need to solve this one is 'two different tangents drawn from an external point to the circle have equal length'

that said, do you have any idea of solving it?

Answer 7

no clue haha

Answer 8

dang is it really much to explain cause this question is hard

Answer 9

...

2 tangents whose legths are the same:

\( 2x^2+3x-1=2x^2-4x+13 \)

\[ 2x^2+3x-1-13=2x^2-4x \]
\[ 2x^2+3x-14=2x^2-4x \]
\[ 2x^2-2x^2+3x-14=-4x \]
\[ 3x-14=-4x \]
\[ 3x+4x=14 \]
\[ 7x=14 \]
\[ x=\frac{14}{7} \]
\[ \LARGE X=? \]

Answer 10

2

Answer 11

is that the answer?

Answer 12

yes.

Answer 13

AB=2? ok

Answer 14

no

Answer 15

x=2

plug that into \( 2x^2+3x−1 \)

Answer 16

oh ok ill solve it give me sec

Answer 17

wait did u plug it in already or i have to and if i have to where do i plug it in

Answer 18

are you serious?

just substitute 2 for x

Answer 19

lmao

Answer 20

i got 3x+7

Answer 21

there are two x-es...

\( 2\cdot 2^2+3\cdot 2−1 \)

solve that

Answer 22

oh i forgot the other one haha

Answer 23

13