Answer:
Area of rectangle possible. : 6ft² ; 10 ft², 12 ft²
Step-by-step explanation:
Feets of fencing = 14
Perimeter = 14
Let x = Length y = width
Perimeter = 2(x + y)
For x = 1
2(1 + y) = 14
1 + y = 7
y =6
Area of rectangle ; x *y = 1 * 6 = 6 ft²
For x = 2
2(2 + y) = 14
2 + y = 7
y = 5
Area of rectangle ; x *y = 2 * 5 = 10 ft²
For x = 3
2(3 + y) = 14
3 + y = 7
y =4
Area of rectangle ; x *y = 3 * 4 = 12 ft²
For x = 4
2(1 + y) = 14
4 + y = 7
y = 3
Area of rectangle ; x *y = 4 * 3 =12 ft²
Hence, Area of rectangle possible. : 6ft² ; 10 ft², 12 ft²
1. 2n + 2*5 =2
2. 2n + 10 = 2
3. 2n = 2 -10
4. 2n= -8
5. n = -8/2
6. n = -4
All points along the circle with be the distance of the radius from the center...so the radius can be found using the Pythagorean Theorem..
r^2=(4-1)^2+(6-2)^2
r^2=9+16
r^2=25
r=5
The equation of the circle can be expressed as:
r^2=(x-h)^2+(y-k)^2 where (h,k) correspond to the center of the circle, (2,1) in this case.
(x-2)^2+(y-1)^2=25
if you wanted it in a more standard form...
(y-1)^2=25-(x-2)^2
(y-1)^2=25-x^2+4x-4
(y-1)^2=-x^2+4x+21
y-1=(-x^2+4x+21)^(1/2)
y=1(+/-)(-x^2+4x+21)^(1/2)
Parallelogram
- quadrilateral (four sided figure)
- oposite sides parallel
<span>Rectangle
</span>- quadrilateral
- equiangular
<span>- right angle (90</span>° angle)<span>
Isosceles Trapezoid
</span>- quadrilateral
- pair of parallel sides
- called "trampezium" in the UK
<span>Square
</span>- regular quadrilateral
- equiangular
- right angles (90°)
PLease state ure question if the bee is on a wall.Whaat size is the bee?
