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2 years ago

What is 1 divided by 5/2

Mathematics

2 answers:

Hunter-Best [27]2 years ago

3 0

The answer is 0.4 or $\frac{2}{5}$

Mars2501 [29]2 years ago

3 0

$\frac{1}{\frac{5}{2} }$

You multiply $\frac{2}{5}$ to the numerator and the denominator to get rid of the fraction in the denominator.

$\frac{1}{\frac{5}{2} }$ × $\frac{\frac{2}{5} }{\frac{2}{5} }$

= $\frac{\frac{2}{5} }{1} =\frac{2}{5}$

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2 years ago

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

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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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2 years ago