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Setler [38]
3 years ago
11

Help help help help help

Mathematics
2 answers:
attashe74 [19]3 years ago
7 0

Answer:

i dont now

Step-by-step explanation:

inde man kona ka balo sorry

Dafna1 [17]3 years ago
3 0

Answer:

quadrant 1 and quadrant IV

Step-by-step explanation:

(a)

secθ > 0 in quadrants 1 and IV

sinθ > 0 in quadrants 1 and II

Thus θ is in quadrant I

(b)

cosθ > 0 in quadrants I and IV

cotθ < 0 in quadrants II and IV

Thus θ is in quadrant IV

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I need help ASAP marking brainliest
Thepotemich [5.8K]

Answer: option D -1,0,2

Step-by-step explanation:

2^x-2

When x=0

We have 2^0-2

1-2=-1

when x=1

We have 2^1-2

2-2=0

When x=2

We have 2^2-2

4-2=2

6 0
3 years ago
Read 2 more answers
1-What is the sum of the series? ​∑j=152j​ Enter your answer in the box.
tangare [24]

Answer:

Please see the Step-by-step explanation for the answers

Step-by-step explanation:

1)

∑\left \ {{5} \atop {j=1}} \right. 2j

The sum of series from j=1 to j=5 is:

∑ = 2(1) + 2(2) + 2(3) + 2(4) + 2(5)

  =  2 + 4 + 6 + 8 + 10

∑ = 30

2)

This question is not given clearly so i assume the following series that will give you an idea how to solve this:

∑\left \ {{4} \atop {k=1}} \right. 2k²

The sum of series from k=1 to j=4 is:

∑ = 2(1)² + 2(2)² + 2(3)² + 2(4)²

  = 2(1) + 2(4) + 2(9) + 2(16)

  =  2 + 8 + 18 + 32

∑ = 60

∑\left \ {{4} \atop {k=1}} \right. (2k)²

∑ = (2*1)² + (2*2)² + (2*3)² + (2*4)²

  = (2)² + (4)² + (6)² + (8)²

  = 4 + 16 + 36 + 64

∑ = 120

∑\left \ {{4} \atop {k=1}} \right. (2k)²- 4

∑ = (2*1)²-4 + (2*2)²-4 + (2*3)²-4 + (2*4)²-4

  = (2)²-4 + (4)²-4 + (6)²-4 + (8)²-4

  = (4-4) + (16-4) + (36-4) + (64-4)

  = 0 + 12 + 32 + 60

∑ = 104

∑\left \ {{4} \atop {k=1}} \right. 2k²- 4

∑ = 2(1)²-4 + 2(2)²-4 + 2(3)²-4 + 2(4)²-4

  = 2(1)-4 + 2(4)-4 + 2(9)-4 + 2(16)-4

  = (2-4) + (8-4) + (18-4) + (32-4)

  = -2 + 4 + 14 + 28

∑ = 44

3)

∑\left \ {{6} \atop {k=3}} \right. (2k-10)

∑ = (2×3−10) + (2×4−10) + (2×5−10) + (2×6−10)  

  = (6-10) + (8-10) + (10-10) + (12-10)

  = -4 + -2 + 0 + 2  

∑ = -4

4)

1+1/2+1/4+1/8+1/16+1/32+1/64

This is a geometric sequence where first term is 1 and the common ratio is 1/2 So

a = 1

This can be derived as

1/2/1 = 1/2 * 1 = 1/2

1/4/1/2 = 1/4 * 2/1 = 1/2

1/8/1/4 = 1/8 * 4/1  = 1/2

1/16/1/8 = 1/16 * 8/1  = 1/2

1/32/1/16 = 1/32 * 16/1  = 1/2

1/64/1/32 = 1/64 * 32/1  = 1/2

Hence the common ratio is r = 1/2

So n-th term is:

ar^{n-1} = 1(\frac{1}{2})^{n-1}

So the answer that represents the series in sigma notation is:

∑\left \ {{7} \atop {j=1}} \right. (\frac{1}{2})^{j-1}

5)

−3+(−1)+1+3+5

This is an arithmetic sequence where the first term is -3 and the common difference is 2. So  

a = 1

This can be derived as

-1 - (-3) = -1 + 3 = 2

1 - (-1) = 1 + 1 = 2

3 - 1 = 2

5 - 3 = 2

Hence the common difference d = 2

The nth term is:

a + (n - 1) d

= -3 + (n−1)2

= -3 + 2(n−1)

= -3 + 2n - 2

= 2n - 5

So the answer that represents the series in sigma notation is:

∑\left \ {{5} \atop {j=1}} \right. (2j−5)

6 0
4 years ago
Perimeter of a semicircle if it’s straight length is 10mm
Ksenya-84 [330]

Here's your answer, I hope you understand this. 30.9cm

6 0
3 years ago
So I keep getting 188 or 88 but I got the reminder right I’m not sure what I did wrong
9966 [12]

Answer:

Step-by-step explanation:

7619 / 7

= 1088 remainder 3.

3 0
4 years ago
I need help asap pls
meriva

Answer:

\displaystyle 36° = m∠NQS

Step-by-step explanation:

Apply the Midsegment Theorem:

\displaystyle \frac{1}{2}[96 - 4x]° = [5x + 6]° → [48 - 2x]° = [5x + 6]° → -42 = -7x; 6 = x \\ \\ [48 - 2(6)]° = [48 - 12]° = 36° \\ \\ AND/OR \\ \\ [5(6) + 6]° = [30 + 6]° = 36°

I am joyous to assist you at any time.

8 0
3 years ago
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