For the triangle do bh\2 and the rectangle do lwh
Answer:
105 ways
Step-by-step explanation:
Because the order of selection does not matter, we know that we will be using a combination instead of a permutation.
₁₅C₂ = 15! / (2! * 13!)
= 15 * 14 * ... * 2 * 1 / 2 * 1 * 13 * 12 * ... * 2 * 1
= 15 * 14 / 2 * 1 (The 13! cancels out)
= 105
Only Statement 2 is surely correct.
because there maybe chances that the line L1 and L3 lies above the line L2 and they can also fulfill the condition of perpendicularity so we can't be sure about statement 3 & statement 1 is clearly incorrect
Answer:
x = -7
y = 1
Step-by-step explanation:
Solve by Substitution :
// Solve equation [2] for the variable x
[2] 7x = 3y - 52
[2] x = 3y/7 - 52/7
// Plug this in for variable x in equation [1]
[1] 4•(3y/7-52/7) + 3y = -25
[1] 33y/7 = 33/7
[1] 33y = 33
// Solve equation [1] for the variable y
[1] 33y = 33
[1] y = 1
// By now we know this much :
x = 3y/7-52/7
y = 1
// Use the y value to solve for x
x = (3/7)(1)-52/7 = -7
Solution :
{x,y} = {-7,1}
