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Stels [109]
2 years ago
13

Decrease 20 sec by 20%

Mathematics
1 answer:
goldfiish [28.3K]2 years ago
4 0
Answer: 16 seconds
To decrease it by 20% percent,
We need to first find amount by:
=20*100/20
=4
Therefore 20% of 20 second is 4 seconds
Now let’s decrease it:
20-4
16
Therefore we get 16 second by decreasing 20% of 20 seconds


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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
3 years ago
Factor 2m^3-12m^2+18m completely
Nina [5.8K]

Answer:

The answer is A.

Step-by-step explanation:

Firstly, you have to take out the common terms for this expression. In this expression, the common terms ard 2 and m :

2 {m}^{3}  - 12 {m}^{2}  + 18m

= 2( {m}^{3}  - 6 {m}^{2}  + 9m)

= 2m( {m}^{2}  - 6m + 9)

Next you have to factorise the brackets :

{m}^{2}  - 6m + 9

=  {m}^{2}  - 3m - 3m + 9

= m(m - 3) - 3(m - 3)

= (m - 3)(m - 3)

=  {(m - 3)}^{2}

So the final answer is :

2m {(m - 2)}^{2}

6 0
3 years ago
Read 2 more answers
If f(x)=2x and g(x)= 1/x, what is the domain of (f*g) (x)
postnew [5]

Answer:

The domain of (f*g) (x) is the set of all real numbers; ( -∞, ∞)

Step-by-step explanation:

(f*g) (x) simply means we obtain the product of f(x) and g(x). We are given that;

f(x)=2x

g(x)= 1/x

(f*g) (x) = f(x) * g(x)

(f*g) (x) = 2x * 1/x = 2

This is a horizontal line defined everywhere on the real line. The domain of  (f*g) (x) is thus ( -∞, ∞)

7 0
3 years ago
Read 2 more answers
Help me please.....:)
kozerog [31]

Answer:I don’t know if I’m right but imma have to say G

Step-by-step explanation:

I’m not good at math

8 0
3 years ago
Answer both questions please
Anna11 [10]
1.) 37/8



2.) .56 X 80 = 44.8

Hope this helps
8 0
3 years ago
Read 2 more answers
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