Answer:
Step-by-step explanation:
In a geometric series, the successive terms differ by a common ratio which is determined by dividing a term by the preceding term.
The formula for determining the nth term of a geometric progression is expressed as
Tn = ar^(n - 1)
Where
a represents the first term of the sequence.
r represents the common ratio between successive terms in the sequence.
n represents the number of terms in the sequence.
From the seies shown,
a = 28
r = 98/28 = 343/98 = 3.5
The formula representing the nth term of the given sequence would be expressed as
Tn = 28 × (3.5)^(n - 1)
Answer:
Sorry man I don't know this I surely would help you if I knew tho
Answer:
-5
Step-by-step explanation:
because it is less than -4 when it is suppose to be more than-4
First I am going to assume that these are both right triangles based off of look and because it is much easier. Without it you have to use law of sines or law of cosines...
So to find x you must first find y which can be done simply by using the pythagorean theorem. This theorem is defined as the sum of the squared legs is equal to the sum of the hypotenuse or x^2 + y^2 = z^2
If we substitute in the known values 16^2 + y^2 = 20^2 and solve for y we get that y = sqrt(20^2 - 16^2), this then simplifies to y = 12
Finding x is much more annoying, the easiest way I can immediately see is to find the upper angles by doing sin(16/20) and then 90 - sin(16/20) since the complementary angle is the one you want. I don't have a calculator or a trig table with me right now but I will tell you that x will be equal to 12 ÷ the inverse cosine of the angle (90degrees - sin(16/20)).
I am pretty sure the answer is D though because we know for sure y = 12 and x has to be greater than y because the hypotenuse must be larger than both legs. It could be E but you won't know unless you do the math for x. So it is either D or E but I would be surprised if a Professor made you do all of the work just to say it doesn't work...
Answer:
12/5
Step-by-step explanation:
All you do is multiply
