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17 Points! Help Me please!

17 Points! Help Me please!-example-1
User Ulisses
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1 Answer

1 vote

Answer:

This is the completed table:
\left[\begin{array}{ccccccccccc}trapezoids&1&2&3&4&5&6&10&18&n\\perimeter&10&16&22&28&34&40&64&112&6n+4\end{array}\right]

for the words and graph please see below and the attached image.

Explanation:

Notice that every time you add a trapezoid in the given fashion (attached to the right of the previous figure) yo are adding a net of 6 units (2+4) to the total perimeter (which started in the first figure as 2+2+2+4 = 10).

We can then write the following values to complete the given table:

1 trapezoid , gives perimeter = 1*4 + 3*2 = 10

2 trapezoids, give perimeter = 2*4 + 4*2 = 16

3 trapezoids, give perimeter = 3*4 + 5*2 = 22

4 trapezoids, give perimeter = 4*4 + 6*2 = 28

5 trapezoids, give perimeter = 5*4 + 7*2 = 34

6 trapezoids, give perimeter = 6*4 + 8*2 = 40

10 trapezoids, give perimeter = 10*4 + 12*2 = 64

n trapezoids, give perimeter = n*4 + (n+2)*2 = 4n + 2n + 4 = 6n + 4

Now, given this general relationship for "n" trapezoids, one can find the number of trapezoids that render a perimeter = 112, as required in the table:

6n + 4 = 112

then 6n = 112 - 4 = 108

then n= 108/6 = 18

so 18 trapezoids give a perimeter of 112 (to complete that missing value in the table).

Now, we can plot a general function of similar form of that we got for a collection of n trapezoids but using "x" instead of "n" and having clear in our mind that the values we want to use are only positive integers greater than zero to represent the number of trapezoids:

f(x) = 6x +4

See attached image were the actual valid points are marked as blue dots.

17 Points! Help Me please!-example-1
User Bharat Soni
by
3.6k points