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3 years ago

What is the distance around a triangle that has sides measuring 21/8 feet, 31/2 feet and 21/2 feet?

1 answer:

5 0

Add all of the sides together. Make sure to use a common denominator because you are adding fractions. Don't forget your unit label (feet)
$\frac{21}{8}$ + $\frac{31}{2}$ + $\frac{21}{2}$
$\frac{21}{8}$ + $\frac{124}{8}$ + $\frac{84}{8}$
$\frac{229}{8}$ feet

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2 years ago

D(x)=(x^2-12x+20)/(3x)

krok68 [10]

Answers:

Vertical asymptote: x = 0

Horizontal asymptote: None

Slant asymptote: (1/3)x - 4

Explanation:

d(x) = $\frac{x^{2}-12x+20}{3x}$

= $\frac{(x-2)(x - 10)}{3x}$

Discontinuities: (terms that cancel out from numerator and denominator):

Nothing cancels so there are NO discontinuities.

Vertical asymptote (denominator cannot equal zero):

3x ≠ 0

÷3 ÷3

x ≠ 0

So asymptote is to be drawn at x = 0

Horizontal asymptote (evaluate degree of numerator and denominator):

degree of numerator (2) > degree of denominator (1)

so there is NO horizontal asymptote but slant (oblique) must be calculated.

Slant (Oblique) Asymptote (divide numerator by denominator):

(1/3)x - 4
3x) x² - 12x + 20
x²
-12x
-12x
20 (stop! because there is no "x")

So, slant asymptote is to be drawn at (1/3)x - 4

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3 years ago