Answer:
A) 4x+8
B) 8x-16
C) 8x+12
D) 6x+6y+6z
Step-by-step explanation:
multiple the number outside of the parentheses by each number or letter inside the parentheses
Answer:
a) x=93
Step-by-step explanation:
Note: Consider the side of first triangle is TQ instead of TA.
Given:
Triangles TQM and TPN which share vertex T.
![TQ\cong TP,TM\cong TN](https://tex.z-dn.net/?f=TQ%5Ccong%20TP%2CTM%5Ccong%20TN)
To find:
The theorem which shows that
.
Solution:
In triangle TQM and TPN,
[Given]
[Given]
[Given]
Since two sides and their including angle are congruent in both triangles, therefore both triangles are congruent by SAS postulate.
[SAS]
Therefore, the correct option is C.
The value of given expression when m = 3 is 27
<h3><u>Solution:</u></h3>
Given expression is ![3m^2](https://tex.z-dn.net/?f=3m%5E2)
We have to evaluate the given expression for m = 3
To find for m is equal to 3, substitute m = 3 in given expression
From given expression,
![\rightarrow 3m^2](https://tex.z-dn.net/?f=%5Crightarrow%203m%5E2)
Plug in m = 3 in above expression
------ eqn 1
We know that,
can be expanded as,
![a^2=a \times a](https://tex.z-dn.net/?f=a%5E2%3Da%20%5Ctimes%20a)
Applying this in eqn 1, we get
![\rightarrow 3(3)^2=3 \times (3 \times 3)](https://tex.z-dn.net/?f=%5Crightarrow%203%283%29%5E2%3D3%20%5Ctimes%20%283%20%5Ctimes%203%29)
Simplify the above expression
![\rightarrow 3(3)^2=3 \times (3 \times 3) = 3 \times 9 = 27](https://tex.z-dn.net/?f=%5Crightarrow%203%283%29%5E2%3D3%20%5Ctimes%20%283%20%5Ctimes%203%29%20%3D%203%20%5Ctimes%209%20%3D%2027)
Therefore, for m = 3 we get,
![\rightarrow 3(3)^2=27](https://tex.z-dn.net/?f=%5Crightarrow%203%283%29%5E2%3D27)
Thus value of given expression when m = 3 is found