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3 years ago

For how many integers $n$ is it true that $\sqrt{n} \le \sqrt{4n - 6} < \sqrt{2n + 5}$?

Mathematics

1 answer:

ale4655 [162]3 years ago

6 0

$\bf \sqrt{n}\leqslant \sqrt{4n-6}$

$\bf \sqrt{n}< \sqrt{2n+5}\implies \stackrel{\textit{squaring both sides}}{n< 2n+5}\implies 0\leqslant 2n - n + 5 \\\\\\ 0 < n+5\implies \boxed{-5 < n} \\\\\\ \stackrel{-5\leqslant n < 2}{\boxed{-5}\rule[0.35em]{10em}{0.25pt}0\rule[0.35em]{3em}{0.25pt}2}$

namely, -5, -4, -3, -2, -1, 0, 1. Excluding "2" because n < 2.

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Ronch [10]

Answer:

25.1cm

Step-by-step explanation:

C=2πr

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2 years ago

Read 2 more answers

12) Write the equation of a rational function

aev [14]

Answer:

$\frac{4(x - 5)(x + 7)(x - 12)}{(x + 1)(x)(x - 12)}$

Step-by-step explanation:

A rational function is

$\frac{p(x)}{q(x)}$

where q(x) doesn't equal zero.

If p is a asymptote, or hole at that value, then we will use

$(x - p)$

Step 1: We have asymptote as 0 and -1 so our denomiator will include

$(x - 0)(x - ( - 1)$

Which is

$(x)(x + 1)$

So our denomator so far is

$\frac{p(x)}{x(x + 1)}$

Step 2: Find Holes.

Since 12 is the value of the hole,

$(x - 12)$

is a the binomial.

This will be both on the numerator and denomator so qe have

$\frac{(x - 12)}{x(x + 1)(x - 12)}$

Step 3: Put the x intercepts in the numerator.

Since 5 and -7 is the intercepts,

$\frac{(x - 12)(x - 5)(x + 7)}{x(x + 1)(x - 12)}$

Step 4: Horinzontal Asymptotes,

Multiply the numerator and denomiator out fully,

$\frac{ {x}^{3} - 10 {x}^{2} - 59x + 420 }{ {x}^{3} - 12 {x}^{2} + x - 12}$

Take a L

look at the coefficients,

Notice they have the same degree,3, this means if we divide the leading coefficents, we will get our horinzonral asymptote.

Multiply the numerator by 4.

$\frac{4(x - 12)(x - 5)(x - 7)}{x(x + 1)(x - 12)}$

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5 0

2 years ago

Factor f(x) = 15x^3 - 15x^2 - 90x completely and determine the exact value(s) of the zero(s) and enter them as a comma separated

Illusion [34]

Answer:

$x=-2,0,3$

Step-by-step explanation:

We have been given a function $f(x)=15x^3-15x^2-90x$ . We are asked to find the zeros of our given function.

To find the zeros of our given function, we will equate our given function by 0 as shown below:

$15x^3-15x^2-90x=0$

Now, we will factor our equation. We can see that all terms of our equation a common factor that is $15x$ .

Upon factoring out $15x$ , we will get:

$15x(x^2-x-6)=0$

Now, we will split the middle term of our equation into parts, whose sum is $-1$ and whose product is $-6$ . We know such two numbers are $-3\text{ and }2$ .