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Scrat [10]
3 years ago
9

Someone please help please I don’t understand this can u please be specific when answering it please help me I’m giving 20points

Mathematics
2 answers:
Len [333]3 years ago
7 0

  Okay so first we need to find 40% of 7.50. To do this you simply divide the percentage by 100, so you have .4 now. You multiply 7.50 by .4 and get 3. Lastly, you just subtract 3 from 7.50 and you get $4.50

Emma will pay $4.50 after the discount.

Aleks04 [339]3 years ago
5 0
So, what you need to do is 7.50 x .40 which is 3...

now minus that from 7.50.

7.50 - 3 = 4.50

So with the stores discount she will actually pay $4.50 for $7.50 worth of candy.
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Help Please! I will mark brainliest!
Damm [24]

Answer:

17

Step-by-step explanation:

4 × 3² = 36

2 × 7 = 14

36 ÷ 12 = 3

3 ± 14 = 17

7 0
3 years ago
There were 276 people on air airplane write a number greater than 276
antoniya [11.8K]
277 good luck on whatever you need this for
7 0
3 years ago
Read 2 more answers
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
Sa se gaseasca numerele a si b a caror medie aritmetica este 200 si care sunt invers proportionale cu 8 si respectiv 4,5.
Olenka [21]
Bonjour, (as you write in roumain)

Soient a et b les 2 nombres

(a+b)/2=200==>a+b=400

8/a=4.5/b==>16b=9a

a+9a/16=400
==> 25a/16=400
==>a=400*16/25
==>a=256
et b= 9*256/16; b=144

3 0
3 years ago
The exterior angle of a certain regular polygon is 60°. How many sides does the polygon have?.
givi [52]
<h3>Answer:  6</h3>

Work Shown:

E = exterior angle = 60 degrees

n = number of sides of the regular polygon

n = 360/E

n = 360/60

n = 6

The regular polygon has 6 sides.

3 0
2 years ago
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