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Katena32 [7]
3 years ago
12

On Earth you weigh 54.5 kilograms, while on the moon you weigh 9,071 grams. How many more grams do you weigh on Earth than on th

e moon?
Mathematics
2 answers:
Nana76 [90]3 years ago
8 0

Answer:

Earth weight:

54.5 kilograms = 54.5 × 10 × 10 × 10 = 54,500 grams

Difference of weights:

54,500 g − 9,071 g = 45,429 grams

Step-by-step explanation:


Sever21 [200]3 years ago
4 0

Answer:

45,429

Step-by-step explanation:

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Can someone help me with these? I need some help.
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Answer:

1) \sqrt{569}  2) 8   3) 0.8  4) \sqrt[10]{61}

Step-by-step explanation:

The equation for pythagoream theorem is

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A = a side length

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To solve these problems plug in the known values into the equation then solve.

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The sum of two numbers is 110. The larger number is 2 less than 7 times the smaller. Find the larger number.
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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)

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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