Which points satisfy the following system of linear inequalities? Show all your work. -5y+4x<18 -15y-9x>117

1 answer:

3 0

The lines intersect on the graph at (-3, -6). The solution set falls within the domain of negative infinity to positive infinity and a range of y is greater than or equal to -6. Without having the capabilities of graphing it to show you, that's the best I can do. Graph the lines (dotted, I presume) and see if you can follow those guidelines for the solution set.

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Find the next two terms of the following sequence: 14, 38, 74, 122, 182, 254...

Bezzdna [24]

Let's find the difference between each number.
38-14=24
74-38=36
122-74=48
182-122=60
254-182=72
As you can tell, you're adding 12 to each difference before. The next difference would be 72+12=84. Let's add.
254+84=338
Now the next difference would be 84+12=96.
338+96=434
So, the next two terms are 338 and 434.

5 0

3 years ago

Dont really understand how to do this

bija089 [108]

Answers:

$t_{10} = -22 \ \text{ and } S_{10} = -85$

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Explanation:

$t_1 = \text{first term} = 5\\t_2 = \text{first term}-3 = t_1 - 3 = 5-3 = 2$

Note we subtract 3 off the previous term (t1) to get the next term (t2). Each new successive term is found this way

$t_3 = t_2 - 3 = 2-3 = -1\\t_4 = t_3 - 3 = -1-3 = -4$

and so on. This process may take a while to reach $t_{10}$

There's a shortcut. The nth term of any arithmetic sequence is

$t_n = t_1+d(n-1)$

We plug in $t_1 = 5 \text{ and } d = -3$ and simplify

$t_n = t_1+d(n-1)\\t_n = 5+(-3)(n-1)\\t_n = 5-3n+3\\t_n = -3n+8$

Then we can plug in various positive whole numbers for n to find the corresponding $t_n$ value. For example, plug in n = 2

$t_n = -3n+8\\t_2 = -3*2+8\\t_2 = -6+8\\t_2 = 2$

which matches with the second term we found earlier. And,

$tn = -3n+8\\t_{10} = -3*10+8\\t_{10} = -30+8\\t_{10} = \boldsymbol{-22} \ \textbf{ is the tenth term}$

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The notation $S_{10}$ refers to the sum of the first ten terms $t_1, t_2, \ldots, t_9, t_{10}$

We could use either the long way or the shortcut above to find all $t_1$ through $t_{10}$ . Then add those values up. Or we can take this shortcut below.

$Sn = \text{sum of the first n terms of an arithmetic sequence}\\S_n = (n/2)*(t_1+t_n)\\S_{10} = (10/2)*(t_1+t_{10})\\S_{10} = (10/2)*(5-22)\\S_{10} = 5*(-17)\\\boldsymbol{S_{10} = -85}$