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pochemuha
3 years ago
8

C^4 ____ c^3 please could you help

Mathematics
1 answer:
Dennis_Churaev [7]3 years ago
8 0

Hello.

Remember the Property of exponents:

\mathrm{\displaystyle\frac{a^{m} }{a^{n} } =a^{m-n}}

Let's use it to solve our maths problem.

\mathrm{\displaystyle\frac{c^{4} }{c^{3} } =c^{4-3} }

Subtract the exponents:

\mathrm{c^{1} =c}

Therefore, the answer is

c.

I hope it helps.

Have a nice day.

\boxed{imperturbability}

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Answer:

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  • The value of b=-25

Step-by-step explanation:

Let

p(x)=6x^4+8x^3-5x^2+ax+b

and

q(x)=2x^2-5

As q(x) is exactly divisible, so, the remainder is 0.

∴ p(x)=q(x).g(x)+r(x)

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As the R.H.S becomes 0 when x = \sqrt{\frac{5}{2} } and x = -\sqrt{\frac{5}{2} }.

And

by substituting x = \sqrt{\frac{5}{2} }

6(\sqrt{\frac{5}{2} } )^4 + 8(\sqrt{\frac{5}{2} } )^3 - 5(\sqrt{\frac{5}{2} } )^2 + a(\sqrt{\frac{5}{2} } ) + b = 0

25 + 20( \frac{\sqrt{5}}{\sqrt{2}} ) + a( \frac{\sqrt{5}}{\sqrt{2}}) +b =0...[A]

by substituting x = -\sqrt{\frac{5}{2} }

25 - 20( \frac{\sqrt{5}}{\sqrt{2}} ) - a( \frac{\sqrt{5}}{\sqrt{2}}) +b =0.....[B]

Adding Equation [A] and equation [B]

50 +2b =0

b=-25

Subtracting Equation [A] and equation [B]

40(\sqrt{\frac{5}{2} } ) + 2a(\sqrt{\frac{5}{2} } ) =0

a = -20

Therefore,

  • The value of a = -20
  • The value of b=-25

<em>Keywords:</em><em> equation, solution</em>

<em> Learn more about equation solution from brainly.com/question/12687078</em>

<em> #learnwithBrainly</em>

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