The width of model is 5.2 inches
<em><u>Solution:</u></em>
Given that, model of space shuttle Apollo Saturn V is created width a scale of 1 : 144
The actual space shuttle has a width of 62.4 feet
Convert 62.4 feet to inches
1 feet = 12 inches
62.4 feet = 12 x 62.4 inches = 748.8 inches
Actual width = 748.8 inches
To find: Width of model in inches
From given,
Scale width : Actual width = 1 : 144
Let "x" be the scaled width
Then we can say,
Scale width : Actual width = 1 : 144
x : 748.8 = 1 : 144
In fractions,
Thus the width of model is 5.2 inches
Answer:
6
Step-by-step explanation:
The ASP of a hexagon is 720 degrees
Answer:
D. 127.5 square units
Step-by-step explanation:
AC=17
BD=15
AB=11.5
Diagonal (1&2) are given, base of the rhombus is also given.
Height is not given
Area of a rhombus given the diagonals
=1/2×d1×d2
Where,
d1=AC=17
d2=BD=15
Area of a rhombus=1/2×d1×d2
=1/2×17*15
=1/2×255
=127.5
D. 127.5 square units
With
we have
so 
has one eigenvalue, 
, with multiplicity 3.
In order for to not be defective, we need the dimension of the eigenspace to match the multiplicity of the repeated eigenvalue 2. But 
has nullspace of dimension 2, since
That is, we can only obtain 2 eigenvectors,
and there is no other. We needed 3 in order to complete the basis of eigenvectors.
Perimiter=legnth+legnth+width+width=2lengt+2width
lesngh=60
perimiter=60+60+2width
perimiter=120+2widht
222=120+2width
subtract 120 from both sdies
102=2width
divide by 2
51 mm=width
