Here is a flower made up of yellow hexagons, red trapezoids, and green triangles. How many copies of this flower pattern could you build if you had 30 yellow hexagons, 50 red trapezoids, and 60 green triangles?

Part a) You could build 5 copies of the flower pattern  

Part b) You would have 40 red trapezoids left over

Step-by-step explanation:

The complete question in the attached figure

Part a)

Let

x -----> the number of yellow hexagons

y ----> the number of red trapezoids

z ----> the number of green triangles

we know that

The flower pattern has the following ratios

Find out how many copies of this flower pattern could you build if you had 30 yellow hexagons,50 red trapezoids, and 60 green triangles

1) For x=30

Divide 30 by 6 (remember that in one pattern there are 6 yellow hexagons)

Verify the quantity of y needed and the quantity of z needed

Find the value of y

10 < 50 ----> is ok

Find the value of z

45<60 --->is ok

2) For y=50

Divide 50 by 2 (remember that in one pattern there are 2 red trapezoids)

Verify the quantity of x needed and the quantity of z needed

Find the value of x

150 > 30 ----> is not ok

3) For z=60

Divide 60 by 9 (remember that in one pattern there are 9 green triangles)

60/9=6.7 copies

Round down

6 copies -----> 6(9)=54 green triangles

Verify the quantity of x needed and the quantity of y needed

Find the value of x

36> 30 --->is not ok

therefore

You could build 5 copies of the flower pattern

Part b)

we know that

x:y:z=6:2:9

If you build 5 copies

1) You would use 5*6=30 yellow hexagons and you would have 0 hexagons left over

2) You would use 5*2=10 red trapezoids and you would have (50-10=40) trapezoids left over

3) You would use 5*9=45 green triangles and you would have (60-45=15) triangles left over

therefore

You would have 40 red trapezoids left over

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.