Consider the following system of equations -2x+5y=19
Y=-5/6x-1/6

Use this graph of the system to approximate its solution

A. (13/4,-5/2)

B. (-5/2,13/4)

C. (-13/4,5/2)

D.) (5/2,-13/4)

C

Step-by-step explanation:

We have the system of equations:

And we want to use the graph to approximate the solution to the system.

Remember that, graphically, the solution to a system of equations is where the two graphs intersect.

We can see that the point at which the two lines intersect is in QII.

Hence, x is negative and y is positive.

The only choices that satisfies these conditions are B or C.

However, for B, -5/2 is -2.5. As we can see from the graph, the x-coordinate of the point of intersection is less than -3.

Hence, the only choice left is C. -13/4 also is -3.25, fulfilling our condition.

So, our final answer is C.

part a

answer: a reasonable domain is 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)

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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

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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).

shell's the art director at her local community center. she paid $10 and for each child that attends she earns $4.

for each child she ears $4 that is the slope m = 4

she paid $10, that is the initial amount paid. so 10 is the y intercept b

m =4   and b = 10

y intercept is 10 so the point is (0,10)

we start graph from y intercept (0,10)

slope = 4 . we know slope = rise / run

slope = 4   or 4/1   are same

rise = 4   and run = 1

for rise we move 4 units up and for run we move 1 unit to the right

the graph is attached below

x axis is the number of child she attends

y axis is the amount   she earns