Which expression is equivalent to the given expression?

Mathematics

1 answer:

Aneli [31]2 years ago

8 0

Answer:

$-\frac{1}{4}(1 + 2y )$

Step-by-step explanation:

Given

$-(\frac{1}{2}y + \frac{1}{4})$

Required

Select equivalent expression

Write each term as a product of 1/4

$-(2 * \frac{1}{4}y + 1 * \frac{1}{4})$

Factorize:

$-\frac{1}{4}(2 * y + 1 * 1)$

$-\frac{1}{4}(2y + 1)$

Apply commutative property:

$-\frac{1}{4}(1 + 2y )$

<em>Hence, option (a) is correct.</em>

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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 $k$ :