X is equal to -11/6 or -1.8333333333..................................
Answer:
27A+81A =108
Step-by-step explanation:
By inspection, it's clear that the sequence must converge to

because

when

is arbitrarily large.
Now, for the limit as

to be equal to

is to say that for any

, there exists some

such that whenever

, it follows that

From this inequality, we get




As we're considering

, we can omit the first inequality.
We can then see that choosing

will guarantee the condition for the limit to exist. We take the ceiling (least integer larger than the given bound) just so that

.
Answer:
if its an arc and center angle is given it could be double the size. IF it is a triangle measure you would use Pythagoras for triangle and trig for given angle, if two triangles are shown and they are scale of each other alternative measures given divide into each measure by the correct line and check this with the matching angles. when found as a division this is the ratio so then you just multiply to find the larger measure but divide to find the smaller measure. AC could also be a junction or vector, if its a type of vector then you just follow the arrows and count how many arrows fit the line pick a direction and ie) if its x2 a then you show a+a.
Step-by-step explanation:
Answer:
x<4
Step-by-step explanation:
3x<12
x<12/3
x<4
I hope this helps