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bulgar [2K]
3 years ago
9

What is the answer to this 2y4 x 5y3

Mathematics
2 answers:
Leni [432]3 years ago
8 0
The answer will be 120y^2
Lubov Fominskaja [6]3 years ago
3 0

Step-by-step explanation:

2 {y}^{4}  \times 5 {y}^{3}  \\  = 10 {y}^{7}

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11. The test scores of 20 students are shown below.
daser333 [38]
Your answer would be 10.6
3 0
2 years ago
Will give brainliest if answer correct// if p(x)= 3x^6 - 5x^5 - x + 7 , then p(2) equals: ??
djverab [1.8K]

\\ \tt\hookrightarrow p(2)=3(2)^6-5(2)^5-2+7

\\ \tt\hookrightarrow p(2)=3(64)-5(32)+5

\\ \tt\hookrightarrow p(2)=192-160+5

\\ \tt\hookrightarrow p(2)=32+5

\\ \tt\hookrightarrow p(2)=37

Option B is correct

4 0
2 years ago
Read 2 more answers
In rectangular form, 6(cos120°, isin120°)
anastassius [24]

Answer:

  (-3, 3√3)

Step-by-step explanation:

Evaluate each of the coordinates. Keep or drop the "i" as your convention requires.

  6(cos(120°), i·sin(120°)) = (6·cos(120°), i·6·sin(120°)) = (6(-0.5), i·6·√3/2)

  = (-3, 3√3 i)

You may want the (x, y) coordinates written as (-3, 3√3).

___

In the complex plane, this is -3+i·3√3.

7 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
What is the perimeter of a triangle with sides of lengths 5, 8, & x?
Anastaziya [24]
The perimeter of a triangle can be found by adding up all the sides

so the perimeter is : 
5 + 8 + x =
13 + x <==
4 0
3 years ago
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