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notsponge [240]

3 years ago

10

How to do this and find the max or min points

2 answers:

Veronika [31]3 years ago

8 0

I see your last line is : c(x) = 0.9(x^2-10)^2 + 101.1

Let y = x^2, then c(y) = 0.9(y-10)^2 + 101.1

Apparently, c(y) is a parabola, min is 101.1 when y = 10, max is infinity

So let x^2 = 10 -> x = sqrt(10) or -sqrt(10), min is 101.1, max is infinity

Levart [38]3 years ago

4 0

Answer:

the min is 101.1

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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]

Answer:

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)= $x^{3} + x+1$ .

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

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7. Quadrilateral MNPQ has vertices M(4,0), N(0,6), P(-4,0) and Q(0, -6). Determine whether

kirill [66]

Answer:

SQUARE

Step-by-step explanation:

If Quadrilateral MNPQ has vertices M(4,0), N(0,6), P(-4,0) and Q(0, -6).

Find the following MN, NP, PQ and MQ

Using the formula for calculating the distance between two points

MN = √(6-0)²+(0-4)²

MN = √6²+4²

MN = √36+16

MN = √52

MN = 2√13

NP = √(0-6)²+(-4-0)²

NP = √6²+4²

NP = √36+16

NP = √52

NP = 2√13

PQ = √(-6-0)²+(0-(-4))²

PQ = √6²+4²

PQ = √36+16

PQ = √52

PQ = 2√13

MQ = √(-6-0)²+(0-4)²

MQ = √6²+4²

MQ = √36+16

MQ= √52

MQ = 2√13

Since the length of all the sides are equal, hence the shape is a SQUARE

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