Ask your own question, for FREE!
Mathematics 7 Online
OpenStudy (mckenzieandjesus):

Suppose you are determining the growth rate of two species of plants. Species A is 12 cm tall and grows 2 cm per month. Species B is 10 cm tall and grows 3 cm per month. Which system of equations models the height of each species H(m) as a function of months m. H(m ) = 12 + 2m H(m ) = 3 + 10m H(m ) = 2 + 12m H(m ) = 3 + 10m H(m ) = 12 + 2m H(m ) = 10 + 3m H(m ) = 2 + 12m H(m ) = 10 + 3m

OpenStudy (mckenzieandjesus):

wouldnt it be a?

OpenStudy (mckenzieandjesus):

@jim_thompson5910 please help?

OpenStudy (kawii2004):

a

OpenStudy (kawii2004):

is correct

OpenStudy (mckenzieandjesus):

are you sure?

OpenStudy (kawii2004):

yes

OpenStudy (kawii2004):

i had the qustion before

jimthompson5910 (jim_thompson5910):

`Species A is 12 cm tall and grows 2 cm per month. ` we start at 12 add 2 cm each month so we have 12 + 2m where m is the number of months

OpenStudy (mckenzieandjesus):

ok

jimthompson5910 (jim_thompson5910):

`Species B is 10 cm tall and grows 3 cm per month` similar to species A, we have 10+3m

OpenStudy (kawii2004):

told you

jimthompson5910 (jim_thompson5910):

H(m ) = 3 + 10m is not correct because this says `start at 3 and increase by 10 each month`

OpenStudy (mckenzieandjesus):

thats what i thought

OpenStudy (mckenzieandjesus):

but the same if it were 2+10..

OpenStudy (mckenzieandjesus):

so it is c

OpenStudy (whpalmer4):

(c) has \[H(m ) = 12 + 2m\]\[ H(m ) = 10 + 3m\] The first equation has a \(y\)-intercept of \(12\) and a slope of \(2\), so it represents a plant with height \(12\) at \(t=0\) and growth of \(2\) per unit of \(t\). The second equation has a \(y\)-intercept of \(10\) and a slope of \(3\), so it represents a plant with height \(10\) at \(t=0\) and a growth of \(3\) per unit of \(t\). These match how the growth of species A and B is described.

OpenStudy (kawii2004):

did you get your awnser

OpenStudy (mckenzieandjesus):

yes thank you :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!