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Anna35 [415]
3 years ago
11

Consider the following geometric sequence. 2, 6, 18, 54,

Mathematics
1 answer:
Kaylis [27]3 years ago
7 0

Answer:

The rule or operation in this sequence is multiplication by 3.

Step-by-step explanation:

2*3=6

6*3=18

18*3=54

and so on...

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DanielleElmas [232]
140 divided by 14 is 10. So x is 10. And Y is 140. 140 divided by y (1) =140
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Answer:

7t=105

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Choose the correct answer to 1,889/36.<br> 42 R17<br> 51 R53<br> 52 R15<br> 52 R17
ozzi

Answer:

52 R17

Step-by-step explanation:

52 * 36 = 1872

1889 - 1872 = 17

--> 52 R17

3 0
3 years ago
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What are the integer solutions to the inequality below?<br> <img src="https://tex.z-dn.net/?f=3%5Cleq%203x-4%5C%20%20%5Ctextless
Rudik [331]

Given:

The compound inequality is:

3\leq 3x-4

To find:

The integer solutions for the given compound inequality.

Solution:

We have,

3\leq 3x-4

Case 1: 3\leq 3x-4

3+4\leq 3x

\dfrac{7}{3}\leq x

2.33...\leq x             ...(i)

Case 2: 3x-4

3x-2x

x                  ...(ii)

Using (i) and (ii), we get

2.33...

The integer values which satisfy this inequality are only 3 and 4.

Therefore, the integer solutions to the given inequality are 3 and 4.

7 0
2 years ago
For questions 13-15, Let Z1=2(cos(pi/5)+i Sin(pi/5)) And Z2=8(cos(7pi/6)+i Sin(7pi/6)). Calculate The Following Keeping Your Ans
weqwewe [10]

Answer:

Step-by-step explanation:

Given the following complex values Z₁=2(cos(π/5)+i Sin(πi/5)) And Z₂=8(cos(7π/6)+i Sin(7π/6)). We are to calculate the following complex numbers;

a) Z₁Z₂ = 2(cos(π/5)+i Sin(πi/5)) * 8(cos(7π/6)+i Sin(7π/6))

Z₁Z₂ = 18 {(cos(π/5)+i Sin(π/5))*(cos(7π/6)+i Sin(7π/6)) }

Z₁Z₂ = 18{cos(π/5)cos(7π/6) + icos(π/5)sin(7π/6)+i Sin(π/5)cos(7π/6)+i²Sin(π/5)Sin(7π/6)) }

since i² = -1

Z₁Z₂ = 18{cos(π/5)cos(7π/6) + icos(π/5)sin(7π/6)+i Sin(π/5)cos(7π/6)-Sin(π/5)Sin(7π/6)) }

Z₁Z₂ = 18{cos(π/5)cos(7π/6) -Sin(π/5)Sin(7π/6) + i(cos(π/5)sin(7π/6)+ Sin(π/5)cos(7π/6)) }

From trigonometry identity, cos(A+B) = cosAcosB - sinAsinB and  sin(A+B) = sinAcosB + cosAsinB

The equation becomes

= 18{cos(π/5+7π/6) + isin(π/5+7π/6)) }

= 18{cos((6π+35π)/30) + isin(6π+35π)/30)) }

= 18{cos((41π)/30) + isin(41π)/30)) }

b) z2 value has already been given in polar form and it is equivalent to 8(cos(7pi/6)+i Sin(7pi/6))

c) for z1/z2 = 2(cos(pi/5)+i Sin(pi/5))/8(cos(7pi/6)+i Sin(7pi/6))

let A = pi/5 and B = 7pi/6

z1/z2 = 2(cos(A)+i Sin(A))/8(cos(B)+i Sin(B))

On rationalizing we will have;

= 2(cos(A)+i Sin(A))/8(cos(B)+i Sin(B)) * 8(cos(B)-i Sin(B))/8(cos(B)-i Sin(B))

= 16{cosAcosB-icosAsinB+isinAcosB-sinAsinB}/64{cos²B+sin²B}

= 16{cosAcosB-sinAsinB-i(cosAsinB-sinAcosB)}/64{cos²B+sin²B}

From trigonometry identity; cos²B+sin²B = 1

= 16{cos(A+ B)-i(sin(A+B)}/64

=  16{cos(pi/5+ 7pi/6)-i(sin(pi/5+7pi/6)}/64

= 16{ (cos 41π/30)-isin(41π/30)}/64

Z1/Z2 = (cos 41π/30)-isin(41π/30)/4

8 0
3 years ago
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