Find the area of the following circles. a. A circle with a 8-inch radius b. A circle with a 10-kilometer radius c. A circle with
2 answers:
Area of a circle is 
. The first one, a, the radius squared is 8*8=64. So 
. If you multiply in pi as 3.14, you'll have 200.96 square inches. The second one, b, the radius squared is 10*10=100. So 
square kilometers. If you multiply in pi as 3.14 you'll have 314 square kilometers. The third one, c, the radius squared is 14*14=196. So 
. Or you could multiply in pi as 3.14 to get 615.44 yd squared. For the last one, d, the radius squared is 22*22=484 cm. Therefore, 
, or multiply in 3.14 for pi to get 1519.76 cm squared. There you go!
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Answer and explanation:
(x + 6)(x + 7)
List the multiples of 42
42: 1 2 3 6 7 14 21 42
Then see which of the multiples add up to 13
1 x 42 = 42 --> 1 + 42 13
2 x 21 = 42 --> 2 + 21 13
3 x 14 = 42 --> 3 + 14 13
6 x 7 = 42 --> 6 + 7 = 13
Answer:
Step-by-step explanation:
Given the series,
∑ ∞ n = 1 − 4 ( − 1 / 2 ) n − 1
I think the series is summation from n = 1 to ∞ of -4(-1/2)^(n-1)
So,
∑ − 4 ( − ½ )^(n − 1). From n = 1 to ∞
There are different types of test to show if a series converges or diverges
So, using Ratio test
Lim n → ∞ (a_n+1 / a_n)
Lim n → ∞ (-4(-1/ 2)^(n+1-1) / -4(-1/2)^(n-1))
Lim n → ∞ ((-4(-1/2)^(n) / -4(-1/2)^(n-1))
Lim n → ∞ (-1/2)ⁿ / (-1/2)^(n-1)
Lim n→ ∞ (-1/2)^(n-n+1)
Lim n→ ∞ (-1/2)^1 = -1/2
Since the limit is less than 0, then, the series converge...
Sum to infinity
Using geometric progression formula
S∞ = a / 1 - r
Where
a is first term
r is common ratio
So, first term is
a_1 = -4(-½)^1-1 = -4(-½)^0 = -4 × 1
a_1 = -4
Common ratio r = a_2 / a_1
a_2 = 4(-½)^2-1 = -4(-½)^1 = -4 × -½ = 2
a_2 = 2
Then,
r = a_2 / a_1 = 2 / -4 = -½
S∞ = -4 / 1--½
S∞ = -4 / 1 + ½
S∞ = -4 / 3/2 = -4 × 2 / 3
S∞ = -8 / 3 = -2⅔
The sum to infinity is -2.67 or -2⅔
Check attachment for better understanding
