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3 years ago

A. Are the expressions 3(m - 2) + 2(m - 2) and 5(m - 2) equivalent expressions?

Mathematics

2 answers:

Lena [83]3 years ago

8 0

Answer:

Step A. 3(m - 2) + 2(m - 2) and 5(m - 2) are equivalent expressions.

Step-by-step explanation:

Step B. If you calculate 3(m - 2) + 2(m - 2) you will see it will equal to 5(m - 2) therefore, they are equal.

Hope I Helped I'm New! :D

Kryger [21]3 years ago

7 0

No, because although when you distribute the 3 and the 2 to m it gets 5m, -6 and -4 =-10 not -2

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Answer:

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Step-by-step explanation:

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2 years ago

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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 $k$ :

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$k \approx \pm 0.028$

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