Answer: absolute value
Step-by-step explanation:
Sorry for my writing it’s horrible ik I’m in a hurry rn lol
Answer:
A
Step-by-step explanation:
I hope this helped you out
Answer:

Step-by-step explanation:
Given in the question an expression,

Step 1
Apply exponential "product rule"





Step 2
Apply exponential " divide rule"




Answer:
every answer except -40 and -20
answer:-77,-100,-45
Treat the matrices on the right side of each equation like you would a constant.
Let 2<em>X</em> + <em>Y</em> = <em>A</em> and 3<em>X</em> - 4<em>Y</em> = <em>B</em>.
Then you can eliminate <em>Y</em> by taking the sum
4<em>A</em> + <em>B</em> = 4 (2<em>X</em> + <em>Y</em>) + (3<em>X</em> - 4<em>Y</em>) = 11<em>X</em>
==> <em>X</em> = (4<em>A</em> + <em>B</em>)/11
Similarly, you can eliminate <em>X</em> by using
-3<em>A</em> + 2<em>B</em> = -3 (2<em>X</em> + <em>Y</em>) + 2 (3<em>X</em> - 4<em>Y</em>) = -11<em>Y</em>
==> <em>Y</em> = (3<em>A</em> - 2<em>B</em>)/11
It follows that

Similarly, you would find

You can solve the second system in the same fashion. You would end up with
