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Fantom [35]
2 years ago
11

WORDED QUADRATIC EQUATION!!! PLS HELP MOSTLY NUMBER 5 PLS

Mathematics
1 answer:
pav-90 [236]2 years ago
3 0

Answer:

7,11

Step-by-step explanation:

The width is x. x+4= the length. The area is side times side, so it’s

x × (x+4) = 77

We can simplify this to:

x × x + x × 4 = 77

x^2 + 4x = 77

x^2 + 4x - 77 = 0 This is a quadratic equation. So we can solve using the quadratic formula:

x = \frac{-4\pm\sqrt{4^2-4\cdot1\cdot-77}}{2\cdot1} ->

\frac{-4\pm18}{2}

\frac{14}{2} \implies7

That’s the positive solution and that’s x, the width. So the width is 7 and the length is 7+4=11

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(1 point) If p(x) and (x) are arbitrary polynomials of degree at most 2, then the mapping =p(-1)q(-1) + p(070) +p(3)q(3) defines
Ainat [17]

If p(x) and q(x) are arbitrary polynomials of degree at most 2 then

||p||||q|| = 26(\sqrt{640}) and angle between p(x) and q(x) is 0.233.

Given that

<p,q> = p(-1)q(-1) + p(0) q(0) + p(3)q(3)

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then the values of p and q at x = -1,0,3 are given as;

x = -1,

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x = 0,

p(0) = 2(0)² + 6 = 6   ,    q(0) = 4(0)² - 4(0) = 0

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p(3) = 2(3)² + 6 = 24  ,  q(3) = 4(3)²- 4(3) = 24.

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         = (8)(8) + (6)(0) + 24(24)

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Now we have to find ||p|| ||q||, for this we'll find ||p|| and ||q||

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     = \sqrt{8(8) + 6(6) + 24(24)}

     = \sqrt{676}

||p|| = 26

and

||q|| = \sqrt{ < q,q > }

      =\sqrt{8(8) + 0(0) + 24(24)}

||q||  =\sqrt{640}

∴||p||||q|| = 26(\sqrt{640\\)

Now we have to find angle between p(x) and q(x),

∴ α = cos⁻¹\frac{ < p,q > }{||p||||q||}

      = cos ⁻¹ \frac{640}{26(\sqrt{640}) }

      = cos ⁻¹ \frac{4\sqrt{10} }{13}

  α  = 13.34°

In radian

α = 0.233.

To know more about Inner product here

brainly.com/question/14185022

#SPJ4

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