Answer:
3
Step-by-step explanation:
15 servings because 11.25/0.75=15
3/4=.75
11 1/4=11.25
15. [4] insufficient information. only given SA
16. There are 3 right triangles, 3 Pythagorean theorem equations
x²+b²= 16²
12²+a² = x²
a²+4² = b²
combine equations using substitution to get in terms of only x
substitute b² for (a² +x²) in first equation
x²+a²+4²=16²
substitute a² for (x² - 12²)
x²+x²-12²+4² = 16²
solve for x
2x² - 144 + 16 = 256
2x² = 256 + 144 - 16
2x² = 384
x² = 384/2
x² = 192
x = 13.85
rounded to nearest tenth, x = 13.9
we can divide the composite shape into on quadrilateral and triangle as shown in the diagram. since the line that joins the two shape is parallel to the base AB, the angle DEC is 85°.The angle CEA is 180°- angle DEC = 180-85=95. Angle BCE is sum of angle CEA, ABC and EAB subtracted from360°= 360°-(95+95+85)=85°. Angle DCE is 120°- ECB=35°. So, angle CDE = 180°-(35°+85°)= 60°. For reference see the diagram.
Answer:
The correct option is;
Use a scale factor of 2
Step-by-step explanation:
The parameters given are;
A = (1, -6)
B = (5, -6)
C = (6, -2)
D = (0, -2)
A'' = (1.5, 4)
B'' = (3.5, 4)
C'' = (4, 2)
D'' = ( 1, 2)
We note that the length of side AB in polygon ABCD = √((5 -1)² + (-6 - (-6))²) = 4
The length of side A''B'' in polygon A''B''C''D'' = √((3.5 -1.5)² + (4 - 4)²) = 2
Which gives;
AB/A''B'' = 4/2 = 2
Similarly;
The length of side BC in polygon ABCD = √((6 -5)² + (-2 - (-6))²) = √17
The length of side B''C'' in polygon A''B''C''D'' = √((4 -3.5)² + (2 - 4)²) = (√17)/2
Also we have;
The length of side CD in polygon ABCD = √((6 -0)² + (-2 - (-2))²) = 6
The length of side C''D'' in polygon A''B''C''D'' = √((4 -1)² + (2 - 2)²) = 3
For the side DA and D''A'', we have;
The length of side DA in polygon ABCD = √((1 -0)² + (-6 - (-2))²) = √17
The length of side D''A'' in polygon A''B''C''D'' = √((1.5 -1)² + (4 - 2)²) = (√17)/2
Therefore the Polygon A B C D can be obtained from polygon A''B''C''D'' by multiplying each side of polygon A''B''C''D'' by 2
The correct option is therefore;
Use a scale factor of 2.
