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3 years ago

14

martin is ordering dvds online. each dvd is priced at $8.50. the wedsite also charges a flat fee of $5.00 for shipping and handl

ing for any size order. if martin decides to spend no more than $56, how many dvds can he buy

1 answer:

Anon25 [30]3 years ago

7 0

Y=8.50x+5.00
56=y
-56=8.50x+5.00

56=8.50x+5.00
-5.00 -5.00
51=8.50x
÷8.5 ÷8.5

6=x <u>HE CAN BUY 6 DVDS</u>

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Answer:

$(x - 5)^2 + (y - 4)^2 + (z - 6)^2 = 14$ .

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Step-by-step explanation:

Apply the Pythagorean Theorem to find the distance between these two endpoints:

$\begin{aligned}&\text{Distance}\cr &= \sqrt{\left(x_2 - x_1\right)^2 + \left(y_2 - y_1\right)^2 + \left(z_2 - z_1\right)^2} \cr &= \sqrt{(6 - 4)^2 + (6 - 2)^2 + ((-3) - (-9))^2 \cr &= \sqrt{56}}\end{aligned}$ .

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$\begin{aligned} r &= \frac{1}{2} \, \sqrt{56} \cr &= \sqrt{\left(\frac{1}{2}\right)^2 \times 56} \cr &= \sqrt{\frac{1}{4} \times 56} \cr &= \sqrt{14} \end{aligned}$ .

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$\begin{aligned}&\left(\frac{x_1 + x_2}{2},\, \frac{y_1 + y_2}{2}, \, \frac{z_1 + z_2}{2}\right) \cr &= \left(\frac{4 + 6}{2},\, \frac{2 + 6}{2}, \, \frac{(-9) + (-3)}{2}\right) \cr &= (5,\, 4\, -6)\end{aligned}$ .

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$\left(x - x_0\right)^2 + \left(y - y_0\right)^2 + \left(z - z_0\right)^2 = r^2$ .

In this case, the equation would be:

$\left(x - 5\right)^2 + \left(y - 4\right)^2 + \left(z - (-6)\right)^2 = \left(\sqrt{56}\right)^2$ .

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$\left(x - 5\right)^2 + \left(y - 4\right)^2 + \left(z + 6\right)^2 = 56$ .

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