Good question, I ask myself the same thibg everyday
Answer:
x=-8
Step-by-step explanation:
I always change the decimals into whole numbers by multiplying them by 10. If you do this, it is much easier to solve. All you do is multiply every term by 10.
the equation changes to
-5(x+3)-2x=41
-5x-15-2x=41
combine like terms
-7x-15=41
bring the 15 over to the 41
-7x=56
divide everything by 7
-x=8
change x to a positive
x=-8
Answer:
tu debes de saber por qué si yo lo sé tú también no es verdad
Let

In order to prove this by induction, we first need to prove the base case, i.e. prove that P(1) is true:

So, the base case is ok. Now, we need to assume
and prove
.
states that

Since we're assuming
, we can substitute the sum of the first n terms with their expression:

Which terminates the proof, since we showed that

as required