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nadezda [96]
3 years ago
10

Which sum is equal to the expression - 5x3 + 2x2 + 9x+2/ (x²+ 2)²

Mathematics
1 answer:
Makovka662 [10]3 years ago
3 0

Answer:

Option C

Step-by-step explanation:

Given expression is \frac{-5x^3+2x^2+9x+2}{(x^2+2)^2}

Option A

\frac{A}{x^2+2}+ \frac{Bx+C}{(x^2+2)^2}

= \frac{A(x^2+2)}{(x^2+2)^2}+ \frac{Bx+C}{(x^2+2)^2}

= \frac{Ax^2+2A+Bx+C}{(x^2+2)^2}

Numerator is a quadratic polynomial, while the expression has the numerator as a cubic .

Therefore, both the expressions are not equivalent.

Option B

\frac{A}{(x^2+2)^2}+ \frac{Bx+C}{(x^2+2)^2}

= \frac{A+Bx+C}{(x^2+2)^2}

Here, numerator of the expression is a linear polynomial which is not equal to given expression.

Therefore, both the expressions are not equivalent.

Option C

\frac{Ax+B}{(x^2+2)}+ \frac{Cx+D}{(x^2+2)^2}

= \frac{(Ax+B)(x^2+2)}{(x^2+2)^2}+ \frac{Cx+D}{(x^2+2)^2}

= \frac{Ax^3+Bx^2+2Ax+2B+Cx+D}{(x^2+2)^2}

= \frac{Ax^3+Bx^2+x(2A+C)+D}{(x^2+2)^2}

Here, numerator is a cubic polynomial.

Therefore, both the expressions are equivalent.

Option D

\frac{Ax+B}{(x^2+2)^2}+ \frac{Cx+D}{(x^2+2)^2}

= \frac{x(A+C)+(B+D)}{(x^2+2)^2}

Numerator is a linear polynomial.

Therefore, sum is not equal to the given expression.

Option C will be the answer.

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Answer:

The ship S is at 10.05 km to coastguard P, and 12.70 km to coastguard Q.

Step-by-step explanation:

Let the distance of the ship to coastguard P be represented by x, and its distance to coastguard Q be represented by y.

But,

<P = 048°

<Q = 360^{o} - 324^{o}

     = 036^{o}

Sum of angles in a triangle = 180^{o}

<P + <Q + <S = 180^{o}

048° + 036^{o} + <S = 180^{o}

84^{o} + <S = 180^{o}

<S  = 180^{o} -  84^{o}

    = 96^{o}

<S = 96^{o}

Applying the Sine rule,

\frac{y}{Sin P} = \frac{x}{Sin Q} = \frac{z}{Sin S}

\frac{y}{Sin P} = \frac{z}{Sin S}

\frac{y}{Sin 48^{o} } = \frac{17}{Sin 96^{o} }

\frac{y}{0.74314} = \frac{17}{0.99452}

⇒ y = \frac{12.63338}{0.99452}

       = 12.703

y = 12.70 km

\frac{x}{Sin Q} = \frac{z}{Sin S}

\frac{x}{Sin 36^{o} } = \frac{17}{Sin 96^{o} }

\frac{x}{0.58779} = \frac{17}{0.99452}

⇒ x = \frac{9.992430}{0.99452}

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x = 10.05 km

Thus,

the ship S is at a distance of 10.05 km to coastguard P, and 12.70 km to coastguard Q.

6 0
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