dentify the vertical asymptotes of f(x) = quantity x plus 6 over quantity x squared minus 9x plus 18.

1 answer:

8 0

So the function we're dealing with is

f(x) = $\frac{x+6}{x^{2}-9x+18 }$

Factorising we get, $x^{2} -9x+18=0$
$(x-3)(x-6)=0$

Therefore, the vertical asymptotes are at x = 3 and x = 6

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Make a number that has a 6 that is 10 times the value of the digit 6 in 165,438 Comparing digits in whole numbers

Ipatiy [6.2K]

Answer:

An example of such number is 1,645,378 ( The value of 6 here is 600,000)

Step-by-step explanation:

Here, we want to compare digits in whole

numbers.

The value of 6 in 165,438 = 60,000

Now, we want a number that is 10 times the value of 6 here

10 times the value of 6 here will be 10 * 60,000 = 600,000

So we want a number which has the value of 6 in it as 600,000

The number in this case as an example can be ;

1,645,378

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3 years ago

6 2/3 times 2 plus 3 1/5 times 2

kramer

Answer:

$19\frac{11}{15}$

Step-by-step explanation:

We have the problem:

$6\frac{2}{3} *2+3\frac{1}{5} *2$

6 2/3 can be written as (6 + 2/3), and 3 1/5 can be written as (3 + 1/5). So, we can then use the distributive property:

$(6+2/3)*2+(3+1/5)*2=12+4/3+6+2/5=18+4/3+2/5$

Now, we need to combine the fractions by finding a common denominator. Here, the common denominator is 3 * 5 = 15: $4/3 +2/5=20/15+6/15=26/15=1\frac{11}{15}$