Answer:
The amount after 4 years = $ 16198.87
Step-by-step explanation:
Points to remember
Compound interest
A = P[1 + R/n]^nt
Were A - Amount
P - Principle
R - Rate of interest
t - Number of years
n - Number of times compounded
<u>To find the amount</u>
Here P = $11,800, R = 8% = 0.08, t = 4 years and n = 4 times
A = P[1 + R/n]^nt
= 11800[1 + 0.08/4]^(4 * 4)
= 16198.87
Therefore amount after 4 years = $ 16198.87
Which of the following is an arithmetic sequence? A.-2, 4, -6, 8, ... B.2, 4, 8, 16, ... C.-8, -6, -4, -2, ...
castortr0y [4]
Answer:
C. -8, -6, -4, -2, ...
Step-by-step explanation:
An arithmetic sequence increases by the same amount every time through addition or subtraction. There is a common difference.
A: -2, 4, -6, 8, ... If there were a common difference, the numbers would not switch between being positive and back to negative. The numbers would either keep going positive or keep going negative.
B: 2, 4, 8, 16, ... The common difference between 16 and 8 is 16 - 8 = 8. The difference between 8 and 4 is 8 - 4 = 4. Since the difference changes between the numbers, this is not an arithmetic sequence.
C. -8, -6, -4, -2, ... The common difference between -2 and -4 is -2 - (-4) = -2 + 4 = 2. The difference between -4 and -6 is -4 - (-6) = -4 + 6 = 2. The difference between -6 and -8 is -6 - (-8) = -6 + 8 = 2. Since the common difference is always two, this is an arithmetic sequence.
Hope this helps!
Answer:
2. 10
3. D. 1 for all n
Step-by-step explanation:
2. The applicable rules of exponents are ...
(a^b)(a^c) = a^(b+c)
a^b = 1/a^-b
__
The value of n is 10.
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3. Using the above rules of exponents, the expression simplifies to ...
6^(-n+n) = 6^0 = 1
The value is 1 for any n.
Answer:
C
Step-by-step explanation:
The center of inscribed circle into triangle is point of intersection of all interior angles of triangle.
The center of circumscribed circle over triabgle is point of intersection of perpendicular bisectors to the sides.
Circumscribed circle always passes through the vertices of the triangle.
Inscribed circle is always tangent to the triangle's sides.
In your case angles' bisectors and perpendicular bisectors intesect at one point, so point A is the center of inscribed circle and the center of corcumsribed circle. Thus, these circles pass through the points X, Y, Z and G, E, F, respectively.
