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shtirl [24]
2 years ago
8

Determine whether the ordered pair (-5, 3) is a solution of the inquality y < 5x + 6.

Mathematics
2 answers:
AnnZ [28]2 years ago
8 0

Answer:

It is not.

Step-by-step explanation:

If you plug in (-5,3) for x and y, you'll get

3 < 5(-5)+6\\3 < -25+6\\3 < -19

That is incorrect. So it is not a solution

Vera_Pavlovna [14]2 years ago
3 0
No, it is not a solution.
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3 years ago
Please see attachment
Dafna11 [192]

Answer:

a) The value of absolute minimum value = - 0.3536  

b) which is attained at   x = \frac{1}{\sqrt{2} }  

Step-by-step explanation:

<u>Step(i)</u>:-

Given function

                       f(x) = \frac{-x}{2x^{2} +1}     ...(i)

Differentiating equation (i) with respective to 'x'

                     f^{l} = \frac{2x^{2} +1(-1) - (-x) (4x)}{(2x^{2}+1)^{2}  }   ...(ii)

                    f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2}  }

Equating Zero

                   f^{l}(x) = \frac{2x^{2}-1}{(2x^{2}+1)^{2}  } = 0

                 \frac{2x^{2}-1}{(2x^{2}+1)^{2}  } = 0

                2 x^{2}-1 = 0

               2 x^{2} = 1

             x^{2}  = \frac{1}{2}

             x = \frac{-1}{\sqrt{2} }  , x = \frac{1}{\sqrt{2} }

<u><em>Step(ii):</em></u>-

Again Differentiating equation (ii) with respective to 'x'

f^{ll}(x) = \frac{(2x^{2} +1)^{2} (4x) - 2(2x^{2} +1) (4x)(2x^{2}-1) }{(2x^{2}+1)^{4}  }

put

      x = \frac{1}{\sqrt{2} }

f^{ll} (x) > 0

The absolute minimum value at   x = \frac{1}{\sqrt{2} }

<u><em>Step(iii):</em></u>-

The value of absolute minimum value

                         f(x) = \frac{-x}{2x^{2} +1}

                       f(\frac{1}{\sqrt{2} } ) = \frac{-\frac{1}{\sqrt{2} } }{2(\frac{1}{\sqrt{2} } )^{2} +1}

         on calculation we get

The value of absolute minimum value = - 0.3536      

<u><em>Final answer</em></u>:-

a) The value of absolute minimum value = - 0.3536  

b) which is attained at   x = \frac{1}{\sqrt{2} }    

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Convert 3,250 mL to L.<br><br> 3.25 L<br> 0.325 L<br> 3,250,000 L<br> 325,000 L
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Answer:

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Answer:

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Triangle:

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Total: 40 + 12 = 52

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Answer:

I'm not to sure but I think it's A, 38.4 mph. By the way are you from Bay doing this for spring break math hw?

Step-by-step explanation:

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6 0
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