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3 years ago

H(x)= (x^2 -36)/x +6 explain why the following function is not continuous at x=-6

Mathematics

1 answer:

sergeinik [125]3 years ago

8 0

You can solve this problem through factoring.

First, you have the equation,

$h(x) = \frac{x^2-36}{x-6}$

Then, you can factor the numerator.

$h(x) = \frac{(x+6)(x-6)}{x-6}$

You can cancel out the x-6 in both the numerator and the denominator because they would equal to just 1.

You are left with $h(x) = x+6$

The function is removable noncontinuous at x=6 because if you plug in 6 in x-6, your denominator would be undefined.

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3 years ago

Find the roots of h(t) = (139kt)^2 − 69t + 80

Sonbull [250]

Answer:

The positive value of $k$ will result in exactly one real root is approximately 0.028.

Step-by-step explanation:

Let $h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ , roots are those values of $t$ so that $h(t) = 0$ . That is:

$19321\cdot k^{2}\cdot t^{2}-69\cdot t + 80=0$ (1)

Roots are determined analytically by the Quadratic Formula:

$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

The smaller root is $t = \frac{69}{38642} - \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ , and the larger root is $t = \frac{69}{38642} + \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$ .

$h(t) = 19321\cdot k^{2}\cdot t^{2}-69\cdot t +80$ has one real root when $\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} = 0$ . Then, we solve the discriminant for $k$ :

$\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}$

$k \approx \pm 0.028$

The positive value of $k$ will result in exactly one real root is approximately 0.028.

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3 years ago

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If the 7/8 pizza is shared between three people, then you have to divide 7/8 by 3. We know that dividing a fraction by 3 is the same as multiplying by 1/3:

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