we know that
if ABC is an isosceles right triangle
then
side AC=side BC
angle A=angle B=45 degrees
cos B=adjacent side angle B/hypotenuse
adjacent side angle B=BC
hypotenuse=AB------> 18 ft
angle B=45 degrees
cos 45°=(√2)/2
so
cos 45°=BC/AB-------> solve for BC
BC=AB*cos 45-------> BC=18*(√2)/2------> BC=9√2 ft
AC=BC--------> AC=9√2 ft
the answer part 1) is
the exact lengths of the two sides, AC and BC is
AC=9√2 ft
BC=9√2 ft
Part b) Find the Area of triangle ABC
Area=b*h/2-------> AC*BC/2-----> (9√2)*(9√2)/2--------> 81 ft²
the answer part b) is
the area of triangle ABC is equal to 81 ft²
6+6
8+4
9+3
10+2
11+1
Hope this helps!
Answer: C
Step-by-step explanation:
450 is her initial value
225 is her rate of change or constant rate
Answer:
Step-by-step explanation:
Let the length of one side of the square base be x
Let the height of the box by y
Volume of the box V = x²y
Since the box is opened at the top, the total surface area S = x² + 2xy + 2xy
S = x² + 4xy
Given
S = 7500sq in.
Substitute into the formula for calculating the total surface area
7500 = x² + 4xy
Make y the subject of the formula;
7500 - x² = 4xy
y = (7500-x²)/4x
Since V = x²y
V = x² (7500-x²)/4x
V = x(7500-x²)/4
V = 1/4(7500x-x³)
For us to maximize the volume, then dV/dx = 0
dV/dx = 1/4(7500-3x²)
1/4(7500-3x²) = 0
(7500-3x²) = 0
7500 = 3x²
x² = 7500/3
x² = 2500
x = √2500
x = 50in
Since y = (7500-x²)/4x
y = 7500-2500/4(50)
y = 5000/200
y = 25in
Hence the dimensions of the box that will maximize its volume is 50in by 50in by 25in.
The Volume of the box V = 50²*25
V = 2500*25
V= 62,500in³
Hence the maximum volume is 62,500in³