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Brilliant_brown [7]
3 years ago
12

Estimate the limit of x-3/x^2-9

Mathematics
1 answer:
julsineya [31]3 years ago
7 0

Answer:

b

Step-by-step explanation:

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What is the sum of the first 100 terms of the sequence 4,9,14,19, ...?
Nadya [2.5K]

Answer:

25150

Step-by-step explanation:

First, we have to see that this is an arithmetic sequence... since to get the next element we add 5 to it.  (a geometric sequence would be a multiplication, not an addition)

So, we have a, the first term (a = 4), and we have the difference between each term (d = 5), and we want to find the SUM of the first 100 terms.

To do this without spending hours writing them down, we can use this formula:

S = \frac{n}{2} * (2a + (n - 1) * d)

If we plug in our values, we have:

S = \frac{100}{2} * (2 * 4 + (100 - 1) * 5) = 50 * (8 + 99 * 5)

S = 50 * (8 + 495) = 50 * 503 = 25150

8 0
3 years ago
Read 2 more answers
PLZZZ HELPPPPPPPPPPP
Oliga [24]

Answer:

Step-by-step explanation:

29 5x+8+x+4=90

6x=78

x=13

30 3x-7+11x-1=90

14x=98

x=7

31 3x+x+8=180

4x=172

x=43

32 6x-1+5x-17=180

11x=198

x=18

7 0
2 years ago
Need to know the answer to the question
Anastaziya [24]
The area is 1680 ft. squared
8 0
3 years ago
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The average of your first 6 tests is 82. If your test average after your seventh test is an 80.5, what did you score on the seve
mart [117]
So average is  of the first 6 tests is 82   well that means 

(test1 + test2+ test3+ test4 + test5+test6)/6 = 82

so now let do some cross multiplliying 

test1 + test2 + test3 + test 4 + test5 +test6 = 82*6

test1 + test2 + test3 + test 4 + test5 +test6  = 492

now lets see if we can find that 7th test score

(test1 + test2 + test3 + test4 + test5 +test6 + test7)/7 = 80.5

So look we found test1 + test2 + test3 + test4 + test5 +test6 to be 492 so lets substitute. 

(492 + test7 )/7 = 80.5

test7 = (80.5*7)-492 = 71.5 



7 0
3 years ago
Brainliest to who answers :>
rusak2 [61]

Answer:

L is not valid.

Step-by-step explanation:

For any triangle, the sum of the two legs must be greater than the hypotenuse. Otherwise the two legs would not be long enough to touch.

7 0
2 years ago
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