30 treadmills : 36 ellipticals
this could be written in a few ways, but i am going to write it as a fraction. since treadmills come first, you put 30/36.
now that you have your fraction you have to find the GCF (greatest common factor) of the two numbers. the GCF is 6, so you have to divide 30 treadmills by 6, and 36 ellipticals by 6.
30/6=5
36/6=6
5/6 or 5:6 is the simplified ratio of treadmills to ellipticals
comment for any questions!!
Answer: The sequence does not have common ratio.
Step-by-step explanation:
By definition, a sequence with a constant ratio (common ratio) between terms is called a "Geometric sequence". In order to find the common ratio, it is necessary to divide any term in the sequence by the previous term.
In this case, given the sequence:
We get:
Since the ratios between the terms are not constant, we can conclude that <em>the sequence does not have common ratio.</em>
Let’s find some exact values using some well-known triangles. Then we’ll use these exact values to answer the above challenges.
sin 45<span>°: </span>You may recall that an isosceles right triangle with sides of 1 and with hypotenuse of square root of 2 will give you the sine of 45 degrees as half the square root of 2.
sin 30° and sin 60<span>°: </span>An equilateral triangle has all angles measuring 60 degrees and all three sides are equal. For convenience, we choose each side to be length 2. When you bisect an angle, you get 30 degrees and the side opposite is 1/2 of 2, which gives you 1. Using that right triangle, you get exact answers for sine of 30°, and sin 60° which are 1/2 and the square root of 3 over 2 respectively.
Now using the formula for the sine of the sum of 2 angles,
sin(A + B) = sin A cos<span> B</span> + cos A sin B,
we can find the sine of (45° + 30°) to give sine of 75 degrees.
We now find the sine of 36°, by first finding the cos of 36°.
<span>The cosine of 36 degrees can be calculated by using a pentagon.</span>
<span>that is as much as i know about that.</span>
Base case: For 
, the left side is 2 and the right is 
, so the base case holds.
Induction hypothesis: Assume the statement is true for 
, that is
We want to show that this implies truth for 
, that
The first 
terms on the left reduce according to the assumption above, and we can simplify the 
-th term a bit:
so the statement is true for all 
.
What i cn help with isto calm down becaus ether eis nothing there so CALM down
