Question

Please Help

Answer 1

@cherry18

Answer 2

Im here hold on

Answer 3

thats math not english

Answer 4

what's the major axis length?
what's the minor axis length?

Answer 5

idk

Answer 6

well.. take a peek at the graph

Answer 7

i honestly dont really know what they mean

Answer 8

well.. have you covered ellipses yet?

Answer 9

but the lines are at 5 and 6

Answer 10

we have but i didnt get it

Answer 11

hmmm

Answer 12

oh ok

Answer 13

hmm well...

the major axis is horizontal, that is, runs over the x-axis
meaning
the "a" will be under the "x"
thus \(\bf \cfrac{(x-{\color{brown}{ h}})^2}{{\color{purple}{ a}}^2}+\cfrac{(y-{\color{blue}{ k}})^2}{{\color{purple}{ b}}^2}=1\qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})
\\ \quad \\
\cfrac{(x-{\color{brown}{ 0}})^2}{{\color{purple}{ 6}}^2}+\cfrac{(y-{\color{blue}{ 0}})^2}{{\color{purple}{ 5}}^2}=1\qquad center\ ({\color{brown}{ 0}},{\color{blue}{ 0}})\)

Answer 14

im confused

Answer 15

hmmm ok.. where?

Answer 16

all of it can u jus tell me which one it is

Answer 17

well... \(\bf \cfrac{(x-{\color{brown}{ h}})^2}{{\color{purple}{ a}}^2}+\cfrac{(y-{\color{blue}{ k}})^2}{{\color{purple}{ b}}^2}=1\qquad center\ ({\color{brown}{ h}},{\color{blue}{ k}})
\\ \quad \\
\cfrac{(x-{\color{brown}{ 0}})^2}{{\color{purple}{ 6}}^2}+\cfrac{(y-{\color{blue}{ 0}})^2}{{\color{purple}{ 5}}^2}=1\qquad center\ ({\color{brown}{ 0}},{\color{blue}{ 0}})
\\ \quad \\
\cfrac{x^2}{6^2}+\cfrac{y^2}{5^2}=1\implies ?\)

Answer 18

...b?

Answer 19

is it?

Answer 20

i think so

Answer 21

well... now you know which is "a" and "b"
just simplify and you'd see which is it

Answer 22

its b