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3 years ago

Find the smallest value of k such that 3240k is a perfect cube

Mathematics

2 answers:

Zarrin [17]3 years ago

3 0

Answer:

225

Step-by-step explanation:

3240k = 2×2×2×3×3×3×3×5×k

Every factor should be in groups of 3

(3×5)² = k

k = 225

kodGreya [7K]3 years ago

3 0

Answer:

Step-by-step explanation:

Prime factorize 3240

3240 = 2*2*2*3*3*3*3*5

To be a perfect cube we have to multiply it by 5*5*3*3 = 25*9=225

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Step-by-step explanation:

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3 years ago

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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)

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$t = \frac{69\pm \sqrt{4761-6182720\cdot k^{2} }}{38642}$

$t = \frac{69}{38642} \pm \sqrt{\frac{4761}{1493204164}-\frac{80\cdot k^{2}}{19321} }$

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 $k$ :

$\frac{80\cdot k^{2}}{19321} = \frac{4761}{1493204164}$

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2 years ago

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ss7ja [257]

Answer:

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Step-by-step explanation:

"The girl on the fifth floor thinks soccer is boring" implies that on the 5th floor lives the football or baseball fan. However, it cannot be the football fan since it is already known that she lives on 6th.

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