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3 years ago

If the vertices of / are B(–3, 3), C(3, 5), and D(–1, 0), what is the perimeter

1 answer:

4 0

Best is to draw a sketch of the three points.
Next step is to find the distances BC, CD, DB.
The perimeter is the sum of the three distances.

The distances are found using the distance formula:
D=sqrt((y2-y1)^2+(x2-x1)^2)
order of (x1,y1), (x2,y2) is not important.
BC=sqrt((3- -3)^2+(5-3)^2)=sqrt(6^2+2^2)=sqrt(40)=5.324
CD=sqrt((-1-3)^2+(0-5)^2)=sqrt(4^2+5^2)=sqrt(41)=6.403
DB=sqrt((-3- -1)^2+(3-0)^2)=sqrt(2^2+3^2)=sqrt(13)=3.606
So perimeter = BC+CD+DB=16.333 (approximately)

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

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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):

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

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