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2 years ago

4(x + 3) ≤ 0 or x+1>3

Mathematics

1 answer:

Semenov [28]2 years ago

3 0

Answer:

Step-by-step explanation:

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I need help on these 2 problems

KIM [24]

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The answer to 17 is -8. 2+ = 10 - 12 = -2*4 = -8.

The answer to 18 is 28. 4^2 = 16, 7 + 1 = 9 + 3 = 12, 16 + 12 = 28.

Step-by-step explanation:

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3 years ago

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I'LL GIVE BRAINLIST Select all the rates that are unit rates.

Alex Ar [27]

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B and D

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2 years ago

Thirteen more than three times a number is 25.

Sladkaya [172]

Answer:

x = 4

Step-by-step explanation:

Equation: 3x + 13 = 25

Step 1: Subtract 13 from both sides.

3x + 13 = 25

- 13 - 13

3x = 12

Step 2: Divide both sides by 3.

3x = 12

3 = 3

x = 4

7 0

3 years ago

Paul wants a bike that cost $400. If he had saved 60% of this amount, how much had he saved.

enyata [817]

Hey!

To solve this problem, we'll first start by creating an equation.

Our Created Equation :

$\frac{400 x 60}{100}$

Now we'll solve it step by step to find the answer.

So we should begin by multiplying the to portion of our fraction.

400 x 60 = 24,000

Next, we divide our answer by 100, since that is the next step to our equation.

24,000 ÷ 100 = 240

So, our answer is...

Paul has already saved $240.00 of the $400.00 he must pay for a new bike.

Hope this helps!

- Lindsey Frazier ♥

4 0

3 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