Given:
It is given that,
PQ ⊥ PS and
∠QPR = 7x-9
∠RPS = 4x+22
To find the value of ∠QPR.
Formula
As per the given problem PR lies between PQ and PS,
So,
∠QPR+∠RPS = 90°
Now,
Putting the values of ∠RPS and ∠QPR we get,

or, 
or, 
or, 
or, 
Substituting the value of
in ∠QPR we get,
∠QPR = 
or, ∠QPR = 
Hence,
The value of ∠QPR is 40°.
Answer:
a = -3/2
Step-by-step explanation:
3(6a+12)=9
Divide each side by 3
3/3(6a+12)=9/3
6a+12 = 3
Subtract 12 from each side
6a+12-12 = 3-12
6a = -9
Divide by 6
6a/6 = -9/6
a = -3/2
I wonder if you mean to write
in place of
...
If you meant what you wrote, then we have


If you meant to write
(the cube root of 256), then we could go on to have
![\sqrt[3]{256}=\sqrt[3]{16^2}=\sqrt[3]{(4^2)^2}=\sqrt[3]{4^4}=\sqrt[3]{4^3\cdot4}=4\sqrt[3]4](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B256%7D%3D%5Csqrt%5B3%5D%7B16%5E2%7D%3D%5Csqrt%5B3%5D%7B%284%5E2%29%5E2%7D%3D%5Csqrt%5B3%5D%7B4%5E4%7D%3D%5Csqrt%5B3%5D%7B4%5E3%5Ccdot4%7D%3D4%5Csqrt%5B3%5D4)