Note: The height of the room must be 3 m instead of 3 cm because 3 cm is too small and it cannot be the height of a room.
Given:
Perimeter of the floor of a room = 18 metre
Height of the room = 3 metre
To find:
The area of 4 walls of the room.
Solution:
We know that, the area of 4 walls of the room is the curved surface area of the cuboid room.
The curved surface area of the cuboid is
Where, h is height, l is length and b is breadth.
Perimeter of the rectangular base is 2(l+b). So,
Putting the given values, we get
Therefore, the area of 4 walls of the room is 54 sq. metres.
The correct choice is B
x is greater than equals to -4.25
Closing parenthesis is misplaced. Should be:
<span>A = h(b₁ + b₂)/2 </span>
<span>h = 75 ft </span>
<span>b₁ = 125 ft </span>
<span>b₂ = 81 ft </span>
<span>Then it's just plug & grind: </span>
<span>A = 75(125 + 81)/2 ft² = 75·206/2 ft² = 75·103 ft² = (7500 + 225) ft² = 7725 ft² </span>
<span>If you follow that, it will guide you through any other, similar p</span>
Answer:
x = -7/40
, y = -11/20
Step-by-step explanation:
Solve the following system:
{12 x - 2 y = -1 | (equation 1)
4 x + 6 y = -4 | (equation 2)
Subtract 1/3 × (equation 1) from equation 2:
{12 x - 2 y = -1 | (equation 1)
0 x+(20 y)/3 = (-11)/3 | (equation 2)
Multiply equation 2 by 3:
{12 x - 2 y = -1 | (equation 1)
0 x+20 y = -11 | (equation 2)
Divide equation 2 by 20:
{12 x - 2 y = -1 | (equation 1)
0 x+y = (-11)/20 | (equation 2)
Add 2 × (equation 2) to equation 1:
{12 x+0 y = (-21)/10 | (equation 1)
0 x+y = -11/20 | (equation 2)
Divide equation 1 by 12:
{x+0 y = (-7)/40 | (equation 1)
0 x+y = -11/20 | (equation 2)
Collect results:
Answer: {x = -7/40
, y = -11/20
Hayden Grace and Gaytar went to the store. They did not go anymore.
