The first example has students building upon the previous lesson by applying the scale factor to find missing dimensions. This leads into a discussion of whether this method is the most efficient and whether they could find another approach that would be simpler, as demonstrated in Example 2. Guide students to record responses and additional work in their student materials.
§ How can we use the scale factor to write an equation relating the scale drawing lengths to the actual lengths?
!
ú Thescalefactoristheconstantofproportionality,ortheintheequation=or=!oreven=
MP.2 ! whereistheactuallength,isthescaledrawinglength,andisthevalueoftheratioofthe drawing length to the corresponding actual length.
§ How can we use the scale factor to determine the actual measurements?
ú Divideeachdrawinglength,,bythescalefactor,,tofindtheactualmeasurement,x.Thisis
! illustrated by the equation = !.
§ How can we reconsider finding an actual length without dividing?
ú We can let the scale drawing be the first image and the actual picture be the second image. We can calculate the scale factor that relates the given scale drawing length, , to the actual length,. If the actual picture is an enlargement from the scale drawing, then the scale factor is greater than one or
> 1. If the actual picture is a reduction from the scale drawing, then the scale factor is less than one or < 1.
Scaffolding:
A reduction has a scale factor less than 1, and an enlargement has a scale factor greater than 1.
Lesson 18: Computing Actual Lengths from a Scale Drawing.
Answer:
- -6x² - 6 = -7x - 9
- -6x² + 7x - 6 + 9 = 0
- -6x² + 7x + 3 = 0
- 6x² - 7x - 3 = 0
<u>Discriminant:</u>
- D = (-7)² - 4*6*(-3) = 49 + 72 = 121
<u>Since D > 0, there are 2 real solutions:</u>
- x = (- (-7) ±√121 )/12
- x = (7 ± 11)/12
- x = 1.5, x = -1/3
Answer:
C.
Step-by-step explanation:
3 times 2 is 6, 2 times 5 is 10, 5 times 3 is 15. However you need 2 for each since there are 3 sets of two equal faces. So it will be 6+6+10+10+15+15
