<h3>Given</h3>
- room height is x feet
- room length is 3x feet
- room width is 3x feet
- a door 3 ft wide by 7 ft tall
<h3>Find</h3>
- The net area of the wall, excluding the door
<h3>Solution</h3>
The area of the wall, including the door, is the room perimeter multiplied by the height of the room. The room perimeter is the sum of the lengths of the four walls.
... gross wall area = (3x +3x +3x +3x)·x = 12x²
The area of the door is the product of its height and width.
... door area = (7 t)×(3 ft) = 21 ft²
Then the net wall area, exclusive of the door is ...
... net wall area = gross wall area - door area
... net wall area = 12x² -21 . . . . square feet
You start off by multiplying the second bracket by -1 to get rid of the bracket.
So,
(2x3 + x2 - 4x) - 9x3 + 3x2 =
2x3 +x2 - 4x - 9x3 + 3x2 =
(2x3 - 9x3) + (x2 + 3x2) - 4x =
-7x3 + 4x2 - 4x
∠ ABD = 5(2X+1)
∠ DBC = 3X+6
∠ EBC = Y +135/2
∠ ABD and ∠ DBC are linear pairs
∴ ∠ ABD +∠ DBC = 180
∴ 5(2X+1) + 3X+6 =180
solve for x
∴ x = 13
∴∠ ABD = 5(2X+1) = 5(2*13+1) = 135
∠ DBC = 3x+6 = 3*13+6 = 45
∠ ABD and ∠ EBC are vertical angles
∴ ∠ ABD = ∠ EBC = 135
∴ y +135/2 = 135
∴ y = 135/2
The <span>statements that are true:
--------------------------------------</span><span>
C.) x=13
E.)measure of angle EBC =135
F.) angle DBC and angle EBC are linear pairs
</span>
Answer: 5k +7
Step-by-step explanation: -2k + -7k + 5k is -4K and 2+3=5 -4K +5 and 6k +3k= 9k + 2 then u do -4K plus 9k and get 5k next, 5+2
Answer:
Subset is part of set which is obtained from Universal set.
Step-by-step explanation:
Like for example :-
A - ( 1,2,3,4,5,6,7,8)
And B - ( 1,2 , 3)
So we can say B is subset of A.
