Please help and show work if you can I only need 10, 11 and 12
1 answer:
Answer:
Step-by-step explanation:
10). AD = BC [If ABCD is a parallelogram]
3x + 1 = 6x - 4
6x - 3x = 1 + 4
3x = 5
x =
11). If ABCD is a parallelogram,
AB = CD
5x - 8 = 2x + 7
5x - 2x = 7 + 8
3x = 15
x = 5
12). If ABCD is a parallelogram,
AB = CD
AB║CD and BD is a transversal intersecting these parallel lines.
∠ABD ≅ ∠BDC [Alternate interior angles]
4x - 1 = x + 38
4x - x = 38 + 1
3x = 39
x = 13
You might be interested in
Answer:
The equation of the line that passes through the point (-2,7) and is perpendicular to the line x-6y=42 is
Step-by-step explanation:
Given:
Let,
point A( x₁ , y₁) ≡ ( -2 , 7)
To Find:
Equation of Line that passes through the point (-2,7) and is perpendicular to the line x-6y=42=?
Solution:
..................Given
which can be written as
Where m is the slope of the line
∴
On Comparing we get
The Required line is Perpendicular to the above line.
So,
Product of slopes = - 1
Slope of the required line is -6
Equation of a line passing through a points A( x₁ , y₁) and having slope m is given by the formula,
i.e equation in point - slope form
Now on substituting the slope and point A( x₁ , y₁) ≡ ( -2, 7) and slope = -6 we get
The equation of the line that passes through the point (-2,7) and is perpendicular to the line x-6y=42 is