Does anyone know the answer to this?

Mathematics

1 answer:

bezimeni [28]3 years ago

5 0

Hey there!

First, let's find s:

s = (5 + 6 + 7) /2 = 9

Second, let's plug in the numbers into the formula:

A = $\sqrt{9(9-7)(9-5)(9-6)}$

Third, let's simplify the numbers to get one number under the square root:

A = [tex]\sqrt{216}

Fourth, we'll find the square root of 216 to get the area:

A = 14.697

Therefore, your answer is A.

Hope this helps!! :)

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$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$

Notice that $x^3+x^2-4x-4=x^2(x+1)-4(x+1)=(x-2)(x+2)(x+1)$ . Decompose the integrand into partial fractions:

$\dfrac{x^2+x-3}{(x-2)^2(x+2)^2(x+1)^2}$
$=\dfrac1{3(x+1)}-\dfrac{11}{32(x+2)}-\dfrac1{3(x+1)^2}-\dfrac1{16(x+2)^2}+\dfrac1{96(x-2)}+\dfrac1{48(x-2)^2}$

Integrating term-by-term, you get

$\displaystyle\int\frac{x^2+x-3}{(x^3+x^2-4x-4)^2}\,\mathrm dx$
$=-\dfrac1{48(x-2)}+\dfrac1{3(x+1)}+\dfrac1{16(x+2)}+\dfrac1{96}\ln|x-2|+\dfrac13\ln|x+1|-\dfrac{11}{32}\ln|x+2|+C$