Answer:
The image coordinates are D'(8, 5), E'(12, 3) and F'(10,0)
Step-by-step explanation:
Given the vertices of △ DEF are D (2, 5), E (6, 3), and F (4, 0). We have to translate the triangle using the vector (6, 0)
Now, translation by vector (6, 0)
D (2, 5)=D'(2+6, 5+0)=D'(8, 5)
E (6, 3)=E'(6+6, 3+0)=E'(12, 3)
F (4, 0)=F'(4+6, 0+0)=F'(10,0)
Now, we have to label the coordinate of the triangle or to get image of triangle after translation.
Hence, The image coordinates are D'(8, 5), E'(12, 3) and F'(10,0)
Given:
It is given that surface area must be less than 150 cm².
Solution:
The Maximum Volume With Total Surface Area Less than 150 cm² is shown in the table.
From the table, it can be concluded that for r=3.00 cm and h=4.95 cm the surface area will be less than 150 cm² and the volume will be the maximum.
Calculate the volume.
Hence, the required dimensions are r=3.00 cm and h=4.95 cm.
Answer:
Width of the uniform path that surrounds the piece of land = 3 m
Step-by-step explanation:
Let the width of the path that surrounds the piece of land be x.
The situation described is sketched in the attached image to this solution.
The dimension of the Length of the piece of land including the uniform path that surrounds it = (40 + 2x) m
The Breadth of the piece of land including the uniform path that surrounds it = (25 + 2x) m
The area of the piece of land including the uniform path that surrounds it
= (Area of the piece of land) + (Area of the uniform path that surrounds it)
Area of the piece of land = Length × Breadth = 40 × 25 = 1000 m²
Area of the uniform path that surrounds it = 426 m² (Given in the question)
The area of the piece of land including the uniform path that surrounds it = 1000 + 426 = 1426 m²
But the area of the piece of land including the uniform path that surrounds it is also
= (Length of the piece of land including the uniform path that surrounds it) × (Breadth of the piece of land including the uniform path that surrounds it
= (40 + 2x) + (25 + 2x)
= 1000 + 80x + 50x + 4x²
= (4x² + 130x + 1000) m²
Equation these 2 areas
4x² + 130x + 1000 = 1426
4x² + 130x - 426 = 0
Solving the quadratic equation
x = 3 or -35.5
Since the width cannot be negative,
x = 3 m
Hope this Helps!!!
