We have that AB || DC.
By a similar argument used to prove that AEB ≅ CED,we can show that (AED) ≅ CEB by (SAS) . So, ∠CAD ≅ ∠ (ACB) by CPCTC. Therefore, AD || BC by the converse of the (
ALTERNATE INTERIOR ANGLES) theorem. Since both pair of opposite sides are parallel, quadrilateral ABCD is a parallelogram
1. AED
2. SAS
3. ACB
4. ALTERNATE INTERIOR ANGLES
<span>Using the information we have
3x+4=40
Do the same to each side of the equation to eliminate for x.
3x+4=40 Minus 4 from each side
3x=40-4
3x=36
Divide 3 from each side
x=36/3
x=12
AC=3x+4
insert the value of x
3(12)+4=40
AC=40
AD=20</span>
Answer:
Its actually one solution it is solvable and only has one solution
Step-by-step explanation:
For 3, it is an isosceles triangle and it is an acute triangle.
Answer:
Option D
Step-by-step explanation:
A type I error occurs when you reject the null hypothesis when it is actually true.
The null hypothesis in this case is minimum breaking strength is less than or equal to 0.5.
A type one error would be allowing the production process to continue when the true breaking strength is below specifications.
