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2 years ago

8

A miner for gold digs at a constant rate. They were able to dig -90 feet in 6 hours. How long have they been digging if they dug

-225 feet?

1 answer:

Pachacha [2.7K]2 years ago

7 0

Answer:

he will dig for 15 hours

Step-by-step explanation:

if a miner digs —90 feets in 6 hours

then he will dig –225 feet in x hours

—225 feet x 6 hours /—90 feet = 15 hours

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Elan Coil [88]

Answer:

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Step-by-step explanation:

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3 years ago

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Plz help and show work

Helen [10]

Answer:

It is A and D.

Step-by-step explanation:

One answer would be 1.50x + 3600 = 1200, because he sells x many bags for $1.50 each, and add 3600 to it

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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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