Please answer correctly !!!!!!!!!! Will mark brainliest !!!!!!!!!!!!!!!
1 answer:
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To solve this problem we must know that when any two lines intersect , a pair of opposite angles from the figure Will be equal
so that means that
we can subtract twenty from each side
now we can subtract like terms
so we can get the final answer as
so that means that
we can subtract twenty from each side
now we can subtract like terms
so we can get the final answer as
A) 
is continuous, so 
is also continuous. This means if we were to integrate 
, the same constant of integration would apply across its entire domain. Over 
, we have 
. This means that
For to be continuous, we need the limit as 
to match 
. This means we must have
Now, over
, we have 
, so 
, which means 
.
b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,
c) is increasing when 
, and concave upward when 
, i.e. when is also increasing.
We have
over the intervals 
and 
. We can additionally see that 
only on and .
d) Inflection points occur when
, and at such a point, to either side the sign of the second derivative 
changes. We see this happening at 
, for which 
, and to the left of we have decreasing, then increasing along the other side.
We also have along the interval 
, but even if we were to allow an entire interval as a "site of inflection", we can see that to either side, so concavity would not change.
For
Now, over
b) Integrating over [1, 3] is easy; it's just the area of a 2x2 square. So,
c)
We have
d) Inflection points occur when
We also have