Because they are using flat nets they may be able to get more boxes per area of cardboard purchased instead of purchasing the boxes and losing some extra cardboard from a cutting or gluing process. the flat nets may also be cheaper to ship or take less area than boxes shipped whole.
I think your question is missed of key information, allow me to add in and hope it will fit the original one. Please have a look at the attached photo.
My answer:
Scale Drawing Lengths: . . / .
Actual Court Lengths . . .
Scale Factor: inch corresponds to ( ∙ ) inches, or inches, so the scale factor is .
Let = , represent the scale drawing lengths in inches, and represent the actual court lengths in inches. The -values must be converted from feet to inches.
To find actual length:
= =
() = inches, or feet
To find actual width:
= = ( )
= / ∙ /
= inches, or feet
The actual court measures feet by feet. Yes, the lot is big enough for the court Vincent planned. The court will take up the entire width of the lot.
Answer:
is the same as
by co-function identities
Step-by-step explanation:
Remember that complementary angles add up to 90°. The angle that i s complementary to 63° is 27°.
Also recall the co-function identities:
- sin (90° – x) = cos x
- cos (90° – x) = sin x
This means that
.
is the size in wheels on the scale model .
<u>Step-by-step explanation:</u>
Correct Question : Tom has a scale model of his car. The scale factor is 1 : 12. If the actual car has 16-inch wheels, what size are the wheels on the scale model?
We have , The scale factor is 1 : 12. We need to find If the actual car has 16-inch wheels, what size are the wheels on the scale model .Let's find out:
Ratio of size of wheels to actual size of wheels is 1:12 , but actual car has 16-inch wheels So ,
⇒
{ x is size of wheel in scale model }
⇒ 
⇒ 
⇒ 
Therefore ,
is the size in wheels on the scale model .
Answer:
(2×2×2)×(2×2×2)×(2×2×2)×(2×2×2)×-(2×2×2)×(2×2×2)