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vfiekz [6]
3 years ago
12

Can someone help me with this worksheet? Will give all my points.

Mathematics
1 answer:
VashaNatasha [74]3 years ago
7 0

Answer:

Here are the formulas

Cube/Rectangular Prism:

Volume = Length*width*height

Surface area = 2(wl+hl+hw)

Lateral area= Area of vertical faces

Base area = length*width

Regular hexagonal prism:

Volume : (3sqrt3/2)*a^2*h

Surface Area  = 6ah+3sqrt(3)a^2

lateral area: 6ah

base area = 3sqrt(3)s^2/2

Triangular prism

Volume: The volume of a triangular prism can be found by multiplying the base times the height.

Surface area: A triangular prism has three rectangular sides and two triangular faces. To find the area of the rectangular sides, use the formula A = lw, where A = area, l = length, and h = height. To find the area of the triangular faces, use the formula A = 1/2bh, where A = area, b = base, and h = height.

Etc.

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Identify the point of tangency in the circle shown​
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Multiplication in GF(24 ): Compute A(x)B(x) mod P(x) in GF(24 ) using the irreducible polynomial P(x) = x 4 + x + 1, where A(x)
tankabanditka [31]

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Step-by-step explanation:

hello,

i advice you check the question again if it is GF(2^{4}) or GF(24). i believe the question should rather be in this form;

multiplication in GF(2^{4}): Compute A(x)B(x) mod P(x) = x^{4} + x+1, where A(x)=x^{2}+1, and B(x)=x^{3} + x+1.

i will solve the above question and i believe with this you will be able to solve any related problem.

A(x)B(x)=(x^{2} +1) (x^{3}+x+1) mod (x^{4}+x+1  ) = (x^{5} +x^{3}+x^{2}  ) + (x^{3}+x+1  ) mod (x^{4} + x+1 )

= x^{5}+2x^{3} +x^{2}  + x + 1 mod(x^{4}+x+1  )

=2x^{2} +1

please note that the division by the modulus above we used

\frac{x^{5}+2x^{3}+x^{2} +1  }{x^{4}+x+1}= x+\frac{2x^{3} +1}{x^{4}+x+1}

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Find each sum. (-3)+2
iren2701 [21]
<h2>(-3)+2 = <u>(</u><u>-1</u><u>)</u></h2>

Hope this will help

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2 years ago
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Help!! Giving brainless need help
charle [14.2K]

Answer:

A. 15

Step-by-step explanation:

V of a right prism which this is. Is LxWxH

So 1.5x 4x2.5=15

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2 years ago
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