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:
x = 110°
Step-by-step explanation:
The Outside Angle Theorem states that the measure of the angle formed by two secants or a secant and tangent from a point outside of a circle is half the difference between the two arcs.
This means that ½ (210 – x) = 50.
½ ( 210 – x ) × 2 = 50 × 2
210 – x = 100.
210 – x + x = 100 + x.
210 = 100 + x.
100 + x = 210.
100 + x – 100 = 210 – 100.
x = 110.
This value must be true because:
½ ( 210 – 110 ) = 50.
½ ( 100 ) = 50.
50 = 50.
Answer: 11/16
<u>Turn 1 7/8 into an Improper Fraction</u>
Multiply 1×8 and get 8. Now add 7 and get 15.
1 7/8 = 15/8
<u>Turn 2 8/11 into an Improper Fraction</u>
Multiply 2×11 and get 22. Now add 8 and get 30.
2 8/11 = 30/11
<u>Do Keep Change Flip</u>
Keep: 15/8
Change ÷ into ×
Flip: 30/11 into 11/30
Your New Problem: 15/8×11/30
<u>Multiply</u>
15/8×11/30=165/240
<u>Simplify</u>
165/240÷15/15=11/16
Answer:
A. DEF=EDF
Step-by-step explanation:
This is a 30.91837% decrease, so approximately a 31% decrease from 98 to 67.7.