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Stels [109]
3 years ago
6

The lengths of the sides of a triangle are in the extended ratio 5 ?: 9 ?: 10. The perimeter of the triangle is 48 cm. What are

the lengths of the? sides?
Mathematics
2 answers:
alexandr402 [8]3 years ago
6 0
A=10, B=18, C=20 are the lengths of the sides
Burka [1]3 years ago
5 0

Answer:

The correct answer is: a = 10 cm, b = 18 cm and c = 20 cm

Step-by-step explanation:

We will use triple proportion:

a : b : c = 5 : 9 : 10     from which is:

a = 5 k , b = 9 k and c = 10 k  where k is coefficient of proportion.

Formula for perimeter is:

P = a + b + c

When we replace value for a = 5 k , b = 9 k and c = 10 k in the formula for P we get:

5 k + 9 k + 10 k = 48  => 24 k = 48 => k = 48/24 = 2

Finally we get:

a = 5 · 2 = 10 cm

b = 9 · 2 = 18 cm

c = 10 · 2 = 20 cm

God with you!!!

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a)

\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3

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But before, we can factor the numerator and the denominator:

\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}

Now, we can conclude that the vertical asymptotes are located at:

\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}

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