Help PLEASE!!! Find X for the triangle.

Mathematics

1 answer:

aleksklad [387]2 years ago

8 0

Answer:

Step-by-step explanation:

$Hypotenuse = 8\sqrt{2} \\Adjacent =x\\\alpha =45\\\\Using\: SOHCAHTOA\\Cos \alpha = \frac{Adjacent}{Hypotenuse}\\\\Cos 45 = \frac{x}{8\sqrt{2} } \\\frac{\sqrt{2} }{2} =\frac{x}{8\sqrt{2} }\\ \\Cross \: Multiply\\2x =16\\\frac{2x}{2} =\frac{16}{2} \\\\x =8$

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The quotient of (x^3+3x^2-4x-12)/(x^2+5x+6)

Elis [28]

Answer: The quotient is (x-2).

Step-by-step explanation:

Since we have given that

$f(x)=(x^3+3x^2-4x-12)\\\\and\\\\g(x)=x^2+5x+6\\\\So,\ \frac{\left(x^3+3x^2-4x-12\right)}{\left(x^2+5x+6\right)}$

Now, we have to find the quotient of the above expression.

So, here we go:

$Factorise\ (x^3+3x^2-4x-12)\\\\=\left(x^3+3x^2\right)+\left(-4x-12\right)\\\\=-4\left(x+3\right)+x^2\left(x+3\right)\\\\=\left(x+3\right)\left(x^2-4\right)$

Now, we will divide the above simplest form with g(x):

$\frac{\left(x+3\right)\left(x^2-4\right)}{\left(x+2\right)\left(x+3\right)}\\\\=\frac{x^2-4}{x+2}\\\\=\frac{\left(x+2\right)\left(x-2\right)}{x+2}\ using\ (a^2-b^2)=(a+b)(a-b)\\\\=x-2$