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SVETLANKA909090 [29]
3 years ago
14

Help pls! It’s due soon

Mathematics
2 answers:
Juliette [100K]3 years ago
6 0

Answer:

x1 = -8 and x2 = 9?

Alja [10]3 years ago
6 0

= x - 8 / x = 2 / x + 9

= Let's solve your equation step-by-step.

= x − 8x = 2x + 9

= x + − 8x = 2x + 9

Multiply all terms by x and cancel:

xx + −8 = 2 + 9x

x² −8 = 9x + 2 ( Simplify both sides of the equation )

x² − 8 − (9x + 2) = 9x + 2 − (9x + 2) (Subtract 9x+2 from both sides)

x² − 9x − 10 = 0

( x + 1 )( x − 10 ) = 0 (Factor left side of equation)

x +1 = 0 or x − 10 = 0 (Set factors equal to 0)

x = − 1 or x = 10

Check answers. (Plug them in to make sure they work.)

<h3>x= −1 ( Works in original equation )</h3><h3>x = 10 ( Works in original equation )</h3><h3 />
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Plz help I have to turn this in by midnight!!!
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I think it is d I am not sure thought I am sorry
6 0
3 years ago
Determine whether the relation R on the set of all Web pages is reflexive, symmetric, antisymmetric, and/or transitive, where (a
xxMikexx [17]

Answer:

a) R is reflexive, R is not symmetric, R is not anti-symmetric, R is transitive.

b) R is reflexive, R is symmetric, R is not anti-symmetric, R is not transitive.

c) R is not reflexive, R is symmetric, R is not anti-symmetric, R is not transitive.

Step-by-step explanation:

a)

(a, b) ∈ R if and only if everyone who has visited Web page a has also visited Web page b.

Obviously R <em>is reflexive</em> (aRa)

Everyone who has visited Web page a has also visited Web page a

R <em>is not symmetric</em> (aRb does not imply bRa)

If everyone who has visited Web page a has also visited Web page b does not mean that everyone who has visited Web page b has also visited Web page a

R <em>is not anti-symmetric</em> (aRb and bRa does not imply a=b)

If everyone who has visited Web page a has also visited Web page b and everyone who has visited Web page b has also visited Web page a does not mean the web pages are the same.

R <em>is transitive</em> (aRb and bRc implies aRc)

If everyone who has visited Web page a has also visited Web page b and everyone who has visited Web page b has also visited Web page c implies that everyone who has visited Web page a has also visited Web page c.

b)

(a, b) ∈ R if and only if there are no common links found on both Web page a and Web page b.

R is obviously <em>reflexive</em> (aRa)

R <em>is symmetric </em>(aRb implies bRa)

if there are no common links found on both Web page a and Web page b, then there are no common links found on both Web page b and Web page a.

R <em>is not anti-symmetric</em> (aRb and bRa does not imply a=b)

if there are no common links found on both Web page a and Web page b and there are no common links found on both Web page b and Web page a does not mean a and b are the same web page.

R <em>is not transitive</em> (aRb and bRc does not imply aRc)

Consider for example three web pages a, b and c such that a and c have a common link and b has no external links at all.

Then obviously (a,b)∈R and (b,c)∈R since b has no links, but (a,c)∉R because they have a common link.

c)

(a, b) ∈ R if and only if there is at least one common link on Web page a and Web page b

R <em>is not reflexive </em>

If the web page a does not have any link at all, then a is not related to a.

R <em>is symmetric </em>(aRb implies bRa)

if there is at least one common link found on Web page a and Web page b, then there is at least one common link found on Web page b and Web page a.

R <em>is not anti-symmetric</em> (aRb and bRa does not imply a=b)

if there is at least one common link found on Web page a and Web page b and there is at least one common link found on Web page b and Web page a does not mean the web pages are the same

R <em>is not transitive</em> (aRb and bRc does not imply aRc)

Consider for example three web pages a, b and c such that a has only two links L1 and L2, b has only two links L2 and L3   c has only two links L3 and L4.  

Then (a, b) ∈ R since a and b have the common link L2, (b, c) ∈ R for b and c have the common link L3, but a and c have no common links, therefore (a,c)∉R

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3 years ago
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Tan t /(sec t -cos t)= sin t/(cos t *(1/cos t -cos t))
=sin t/(cos t*(1-cos ²t)/cos t))
=sin t/sin² t=1/sin t [or=cosec t.]
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Isoke is solving the quadratic equation by completing the square.
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3 years ago
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