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zavuch27 [327]
3 years ago
6

Write each ratio statement as a fraction and reduce to lowest terms if possible:

Mathematics
1 answer:
qwelly [4]3 years ago
7 0

The answers are:

18 to 24 =  \frac{18}{24} = \frac{3}{4}

5 to 15 =  \frac{5}{15} = \frac{1}{3}

14:42 =  \frac{14}{42} = \frac{1}{3}

15 cents to 18 cents = \frac{15}{18}  = \frac{5}{6}

Explanation:

To write a fraction using a ratio, simply use the first number as the numerator (top number), in this case, the numbers 18, 5, 14, and 15, and the second or biggest number as a denominator (bottom number), in this case, the numbers 24, 15, 42 and 18. This means the fractions are  \frac{18}{24},  \frac{5}{15}, \frac{14}{42} , and  \frac{15}{18}.

The second step is to reduce or simplify the fractions, which means the numbers in a fraction are divided by the same factor (a number that divides another without a remainder). Additionally, to do this, it is important to reduce the fraction to its minimum.

\frac{18}{24} divide this by 6, which is equivalent to \frac{3}{4}

\frac{5}{15} divide this by 5, which is equivalent to \frac{1}{3}

\frac{14}{42} divide this by 14. which is equivalent to \frac{1}{3}

\frac{15}{18} divide this by 3, which is equivalent to \frac{5}{6}

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Please help!
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a) The polynomial in expanded form is f(x) = x^{3}-2\cdot x^{2}-21\cdot x -18.

b) The slant asymptote is represented by the linear function is y = -x + 1.

c) There is a discontinuity at x = 2  with a slant asymptote.

a) In this question we are going to use the Factor Theorem, which establishes that polynomial are the result of products of binomials of the form x-r_{i}, where r_{i} is the i-th root of the polynomial and the grade is equal to the quantity of roots. Therefore, the polynomial f(x) has the following form:

f(x) = (x-6)\cdot (x+1)\cdot (x+3)

And the expanded form is obtained by some algebraic handling:

f(x) = (x-6)\cdot (x^{2}+4\cdot x +3)

f(x) = x\cdot (x^{2}+4\cdot x + 3)-6\cdot (x^{2}+4\cdot x +3)

f(x) = x^{3}+4\cdot x^{2}+3\cdot x -6\cdot x^{2}-24\cdot x -18

f(x) = x^{3}-2\cdot x^{2}-21\cdot x -18 (1)

The polynomial in expanded form is f(x) = x^{3}-2\cdot x^{2}-21\cdot x -18.

b) In this question we divide the polynomial found in a) (in factor form) by the polynomial x^{2}-x -2 (also in factor form). That is:

g(x) = \frac{(x-6)\cdot (x+1)\cdot (x+3)}{(x-2)\cdot (x+1)}

g(x) = \frac{(x-6)\cdot (x+3)}{x-2} (2)

The slant asymptote is defined by linear function, whose slope (m) and intercept (b) are determined by the following expressions:

m =  \lim_{x \to \pm \infty} \frac{g(x)}{x} (3)

b =  \lim_{x \to \pm \infty} [g(x)-x] (4)

If g(x) = \frac{(x-6)\cdot (x+3)}{x-2}, then the equation of the slant asymptote is:

m =  \lim_{x \to 2} \frac{(x-6)\cdot (x+3)}{x\cdot (x-2)}

m =  \lim_{x \to \pm \infty} \left(\frac{x^{2}-3\cdot x -18}{x^{2}-2\cdot x} \right)

m =  1

b =  \lim_{x \to \pm \infty} \left(\frac{x^{2}-3\cdot x -18}{x-2}-x \right)

b =  \lim_{x \to \pm \infty} \left(\frac{x^{2}-3\cdot x - 18-x^{2}+2\cdot x}{x-2}\right)

b =  \lim_{n \to \infty} \left(\frac{-x-18}{x-2} \right)

b = -1

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We kindly invite to check this question on asymptotes: brainly.com/question/4084552

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