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Svetlanka [38]
3 years ago
13

A proton represents a positive charge. Write an integer to represent 5 protons. An electron represents a negative charge. Write

an integer to represent 3 electrons.
Mathematics
1 answer:
Ugo [173]3 years ago
8 0

Answer

Given,

proton represent positive charge                                    

the integer that will give the  representation of 5 protons will be equal to + 5

an electron represents negative charge                                            

the integer that will give the representation of 3 electrons will be equal to -3

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The perimeter of a rectangular field is 316 yards. If the length of the field is 84 yards, what is its width?
4vir4ik [10]

Answer:

74

Step-by-step explanation:

84+84=168

316-168=148

148/2=74

5 0
3 years ago
Sin (3x)cos(6x) - cos (3x) sin(6x)=-0.9
geniusboy [140]

Answer:

x=21.38602241

Step-by-step explanation:

Given: sin(3x)cos(6x) - cos(3x)sin(6x) = -0.9

Rewrite as difference identity: sin(3x-6x)=-0.9

Combine like terms in parentheses: sin(-3x)=-0.9

Take the inverse sine of both sides: -3x=-64.15806724

Divide both sides by -3: x=21.38602241

Refer to the sheet to see all trigonometric identities

6 0
3 years ago
Read 2 more answers
1161.90 to the nearest hundred
forsale [732]
1 thousand
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8 0
3 years ago
A dollar bill is worth _____ pennies. When written in decimal format, each penny can be written as ______ of a dollar.
Mkey [24]
One dollar is 100 pennies. In decimal for, each penny can be written as .01
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4 0
3 years ago
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Find the quotient of these Complex Numbers. <br><br><br> (5-i) / (3+2i)
JulsSmile [24]

Let the given complex number

z = x + ix = \dfrac{5-i}{3+2i}

We have to find the standard form of complex number.

Solution:

∴ x + iy = \dfrac{5-i}{3+2i}

Rationalising numerator part of complex number, we get

x + iy = \dfrac{5-i}{3+2i}\times \dfrac{3-2i}{3-2i}

⇒ x + iy = \dfrac{(5-i)(3-2i)}{3^2-(2i)^2}

Using the algebraic identity:

(a + b)(a - b) = a^{2} - b^{2}

⇒ x + iy = \dfrac{15-10i-3i+2i^2}{9-4i^2}

⇒ x + iy = \dfrac{15-13i+2(-1)}{9-4(-1)} [ ∵ i^{2} =-1]

⇒ x + iy = \dfrac{15-2-13i}{9+4}

⇒ x + iy = \dfrac{13-13i}{13}

⇒ x + iy = \dfrac{13(1-i)}{13}

⇒ x + iy = 1 - i

Thus, the given complex number in standard form as "1 - i".

5 0
3 years ago
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