Mathematics, 23.10.2019 23:10, ayoismeisjjjjuan

Jelanie wrote the system of linear equations below to represent the number of student and adult tickets he and his friend sold to their school play 12s+10a=128 16s+21a=232 jelani also wrote the augmented sound matrix to represent this system which matrix could melanin have written

Jelanie wrote the system of linear equations below to represent the number of student and adult tick

Answers: 2

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Answers

answered: shelbycade230

$\left[\begin{array}{ccc}12&10&|128\\16&21&|232\end{array}\right]$

Step-by-step explanation:

The two equations Jelani wrote are:

12s+10a=128

16s+21a=232

The coefficient matrix is:

$\left[\begin{array}{cc}12&10\\16&21\end{array}\right]$

The constant matrix is:

$\binom{128}{232}$

The augmented matrix is the combination of the coefficient matrix and the constant matrix.

$\left[\begin{array}{ccc}12&10&|128\\16&21&|232\end{array}\right]$

answered: yair7

$\large\boxed{\left[\begin{array}{ccc}12&10&128\\16&21&232\end{array}\right] }$

Step-by-step explanation:

$\left\{\begin{array}{ccc}ax+by=n\\cx+dy=m\end{array}\right\Rightarrow\left[\begin{array}{ccc}a&b&n\\c&d&m\end{array}\right]$

$\text{We have the system of equations:}\\\\\left\{\begin{array}{ccc}12s+10a=128\\16s+21a=232\end{array}\right\Rightarrow\left[\begin{array}{ccc}12&10&128\\16&21&232\end{array}\right]$

answered: Guest

step-by-step explanation:

answered: Guest

part a

answer: a reasonable domain is $0 \le n \le 10$ where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).