Answer:
y=7
Step-by-step explanation:
The two given angels are congruent so they must be equal in measure.
Set them up equal and find y.
8y = 5y + 21, subtract 5y from both sides
8y -5y = 21, combine like terms
3y = 21, divide both sides by 3
y = 7
2nd one is the correct one I think
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:
centre (5, 6 ) , r = 
Step-by-step explanation:
the equation of a circle in standard form is
(x - h)² + (y - k)² = r²
where (h, k ) are the coordinates of the centre and r the radius
given
x² + y² - 10x - 12y + 24 = 0 ( collect x and y terms together and subtract 24 from both sides )
x² - 10x + y² - 12y = - 24
using the method of completing the square
add ( half the coefficient of the x / y terms)² to both sides
x² + 2(- 5)x + 25 + y² + 2(- 6)y + 36 = - 24 + 25 + 36
(x - 5)² + (y - 6)² = 37 ← in standard form
with centre (h, k ) = (5, 6 ) and r = 
Answer:
A) Subtract 5 from each side
Step-by-step explanation:
This would be your first step, then you would want to isolate your variable so you would divide both sides by 2.