Answer:- AAS postulate
Explanation:-
- AAS postulate tells that if two angles and a non-included side of a triangle to equal to the two angles and a non-included side of another triangle then the two triangles are said to be congruent.
Given:- One angle and one side of a triangle is equal to the one angle and one side of the other triangle.
We see there is one more pair of equal angles as they are vertically opposite angles . [See the attachment]
⇒ there is a triangle where two angles and a non-included side of a triangle to equal to the two angles and a non-included side of another triangle then the two triangles are said to be congruent.
⇒ The triangles are congruent [ by ASA postulate]
Answer: C. y=25x+125
Step-by-step explanation:
it multiplies by 75 and adds 125 each time
Solve for x over the real numbers:18 ° (x - 1) (x - 10 ° x) = 0
Divide both sides by 18 °:(x - 1) (x - 10 ° x) = 0
Split into two equations:x - 1 = 0 or x - 10 ° x = 0
Add 1 to both sides:x = 1 or x - 10 ° x = 0
Collect in terms of x:x = 1 or (1 - 10 °) x = 0
Divide both sides by 1 - 10 °:Answer: x = 1 or x = 0
Answer:
Class Boundary = 1 between the sixth and seventh classes.
Step-by-step explanation:
Lengths (mm) Frequency
1. 140 - 143 1
2. 144 - 147 16
3. 148 - 151 71
4. 152 - 155 108
5. 156 - 159 83
6. 160 - 163 18
7. 164 - 167 3
The class boundary between two classes is the numerical value between the starting value of the higher class, which is 164 for the 7th class in this case, and the ending value of the class of the lower class, which is 163 for the 6th class in this case.
Therefore the class boundary between the sixth and seventh classes
= 164 - 163 = 1
Therefore Class Boundary = 1.
It can be seen that class boundary for the frequency distribution is 1.
If we take the difference between the lower limits of one class and the lower limit of the next class then we will get the class width value.
Therefore, Class width,
w = lower limit of second class - lower limit of first class
= 144 - 140
= 4
