geno3141

Let the constant be x:

By the definition of a geometric sequence, then:  (60 + x)/(100 + x)  =  (100 + x)/(150+ x)

Cross multiplying:  (60 + x)(150 + x)  =  (100 + x)(100 + x)

--->   9000 + 210x + x²  =  10000 + 200x + x²

--->   10x  =   1000

--->   x = 100 

The epression is 1337(-½)ⁿ and you want this > 1

The negative sign flips the answer from negative to positive, to negative, etc.

I'm going to ignore the negative sign, but realize that my answer will be two times too large, because it will include both the positive and the negative answers.

1337(½)ⁿ  >  1   --->   (½)ⁿ  >  1/1337   --->   2ⁿ  <  1337   --->   n = 10

But, this contains both the negative and the positive answers, so, removing the negative solutions,  n = 5.

You can use the sum of an arithmetic series to find the sum of: 3 + 13 + 23 + ... + 93  =  480

Now, to find the sum of the aritmetic series: 103 + 113 + 123 + ... + 193:  there are 10 of these numbers and each number is 100 greater than the number in the previous series, so their sum is 10 x 100 + 480  =  1480.

Similarly, 203 + 213 + 223 + ... + 293  =  1000 + previous answer  =  2480.

...

903 + 913 + 923 + ... + 993  = 9480.

You can use the sum of an arithmetic series to find the sum of these answers:  480 + 1480 + 2480 + ... + 9480.

Now, for those numbers ending in 7:  each of them is 4 more than the corresponding number ending in 3. So, to the answer for those numbers ending in 3, we must add 4 x the number of numbers ending in 7.

Add these two answers together.

Part a:  There are n terms, so the sum of n terms of a is an.

The sum of the n terms: 0, 1, 2, 3, ..., n-1  =  n(0 + n - 1)/2  =  n(n - 1) / 2

The total sum is:  an + n(n - 1)/2

When creating these two elements subsets, you must use each of the elements 6 times (so each can be matched with each of the other 6 numbers).

So you need 6 1's, 6 2's, 6 3's, ..., 6 7's.

6[1 + 2 + 3 + 4 + 5 + 6 + 7]  =  6[28]  =  168

The first number is 1, the last number is 125; their sum is 126, and their mean is 63.

The second number is 2, the second last number is 124; their sum is 126, and their mean is 63.

Etc.

The middle number is 63.

What you are looking at is a sequence of 63s; therefore, their mean is 63.

Do the numbers work in the Pythagorean Theorem?         c²  =  a² + b²

c must be replaced with the largest value; the other two values replace a and b, in either order:

So, is this true:  6²  =  2² + 5²

If so, these create a right triangle; if not, it's not a right triangle.

OK?

Since you have not included the graph, we have no way of knowing. However, you can figure out which one by finding one of the ordered pairs of the graph and putting these values into every equation. (Don't stop when you find the first one that works!).

If only one equation works, you have found it; if more than one equation works, take another point, and put its values into the equations that worked for the first point. Keep doing this until you have only one equation that always works.

It will reduce your work if you realize that the first equation creates a straight line; so if the graph is a straight line, that's your answer.

The last equation creates a "V" shape.

The middle two equations are parabolas -- "U" shapes. 

The commutative property exchange terms, such as  x + 9  becomes  9 + x.  (In the same sense that a commuter moves from one place in a city to anothern place.)

The associative property rearranges parentheses, such as  x + (3x + 9)  becomes  (x + 3x) + 9.  It changes which terms are associated together; in the first place, the  3x  is associated with the  9;  in the second case the  3x  is associated with the  x.

The distributive property either multiplies out or it factors. For example, when  3(x + 7)  becomes 3x + 21, you have multiplied out the problem; when  4x + 20  becomes  4(x + 5),  you have factored the problem. In both cases, written one way it has parentheses, written the other way, it has no parentheses.

Combining like terms simplifies problems;  3x + 7 + 9  becomes  3x + 16  (the numbers are combined)  and    3x + 8x + 12  becomes  11x + 12. The problem becomes shorter, simpler.

In your problem, the  3 + 2m  is rewritten  as  2m + 3, terms are exchanged.

Now, can you figure out which rule is being used?

Again, replace x with each of its values:

x = -2   --->   y  =  -[ (-2)² ]  =  -[ 4 ]  =  -4         --->         (-2, -4)

x = -1   --->   y  =  -[ (-1)² ]  =  -[ 1 ]  =  -1         --->         (-2, -1)

Etc.

Connect these points, in order, with a smooth curve, and extend the curve both to the right and to the left.

OK?

If your equation has no exponents on the x-term, it will create a straight line.

If your equation has an exponent of two (and no other exponents -- except for, maybe, one), it will create a parabola.