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3 years ago

Please help as soon as possible

Mathematics

1 answer:

olga nikolaevna [1]3 years ago

7 0

Answer:

srry but i can't see the graph

Step-by-step explanation:

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0-12/3-0

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3 years ago

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Which equals the product of (x-3)(2x + 1)? 222 – 7x - 3 22 – 5x - 3 3x = 2 6,2

Katena32 [7]

Answer:

Step-by-step explanation:

$2 {x}^{2} - 5x - 3$

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3 years ago

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Margaret [11]

hopefully this helps

Step-by-step explanation:

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2 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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2 years ago

If 1 oz of potato chips contains 180 calories, how many ounces would contain 3,180 calories

Ronch [10]

Do you need to show work?

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3 years ago

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