It’s 90 degrees hope this helps
* parenthesis tell you to multiply whatever is outside them by whatever is inside them, so let’s do that!
All we need is a calculator !!
1. 0.3(p) - 0.3(3)
0.3p - 0.9
2. 1.2(0.3x) - 1.2(1.4)
0.36x - 1.68
3. -3(2x) - 3(5)
-6x - 15
4. -7(2k) - 7(-h)
-14k + 7h
5. -1/2(6x) - 1/2( -1/3)
-3x + 1/6
6. -3(0.2m) - 3(5)
-0.6m - 15
7. -0.6(0.4y) - 0.6(-1)
-.24y + 0.6
* remember that when you multiply to negatives, it equals a positive!!
Hope I helped :)
Answer:
Option A
![m\angle LQR=126^o](https://tex.z-dn.net/?f=m%5Cangle%20LQR%3D126%5Eo)
Step-by-step explanation:
step 1
Find the value of b
we know that
----> by vertical angles
substitute the given values
![(-3b+63)^o=(90-12b)^o](https://tex.z-dn.net/?f=%28-3b%2B63%29%5Eo%3D%2890-12b%29%5Eo)
solve for b
![12b-3b=90-63\\9b=27\\b=3](https://tex.z-dn.net/?f=12b-3b%3D90-63%5C%5C9b%3D27%5C%5Cb%3D3)
step 2
Find the measure of angle LQR
we know that
---> by supplementary angles (form a linear pair)
![m\angle MQR=(-3b+63)^o](https://tex.z-dn.net/?f=m%5Cangle%20MQR%3D%28-3b%2B63%29%5Eo)
substitute the value of b
![m\angle MQR=(-3(3)+63)=54^o](https://tex.z-dn.net/?f=m%5Cangle%20MQR%3D%28-3%283%29%2B63%29%3D54%5Eo)
substitute in the expression above
![m\angle LQR+54^o=180^o](https://tex.z-dn.net/?f=m%5Cangle%20LQR%2B54%5Eo%3D180%5Eo)
![m\angle LQR=180^o-54^o=126^o](https://tex.z-dn.net/?f=m%5Cangle%20LQR%3D180%5Eo-54%5Eo%3D126%5Eo)