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Nimfa-mama [501]

3 years ago

11

It took researchers three weeks analyze a large amount of data using a super computer how many minutes did it take them to analy

ze the data

1 answer:

Helen [10]3 years ago

3 0

3 weeks = 21 days = 504 hours = 30,340 minutes

hope that helped

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The pen for Ruth’s dog is 7 feet long and 6 1/2 feet wide. Her friend Ralph’s dog pen is 14 feet long and 6 1/2 feet wide. How m

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Answer:

2 times larger

Step-by-step explanation:

Ralph's pen is 91 feet long

ruths is 45.5

Easily makes 91

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2 years ago

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Given lines p and q are parallel, find the value of x that makes each diagram true.

Klio2033 [76]

Answer: for a the answer is 140° and for b x=25°

Step-by-step explanation:

for a, when you have 2 parallel lines cut by a transversal,the corresponding angles are congruent. and the angle140 is congruent to the corresponding angle as they are vertical angles are congruent.

for b, angle x is congruent to the angle that is supplementary to 155 therefore 180-155=25°

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3 years ago

Brent is making a playlist for an upcoming trip. The average length of a song is 3.53 minutes If the flight is about 120 minutes

vlada-n [284]

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34 songs

Step-by-step explanation:

120min/3.53 = 33.99

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3 years ago

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I Need help with 1-5

natta225 [31]

Answer:

1)=0

2)

$\frac{ \sqrt{3} }{?2}$

3)

$\sqrt{3}$

4)

$2 + \sqrt{3}$

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2 years ago

A Cepheid variable star is a star whose brightness alternately increases and decreases. For a certain star, the interval between

sattari [20]

Answer:

a)

$B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)$

b) 0.09

Step-by-step explanation:

We are given the following in the question:

$B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)$

where B(t) gives the brightness of the star at time t, where t is measured in days.

a) rate of change of the brightness after t days.

$B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(t) = 0.45\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\times \dfrac{2\pi}{4.4}\\\\B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)$

b) rate of increase after one day.

We put t = 1

$B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(1) = \dfrac{0.9\pi}{4.4}\bigg(\cos(\dfrac{2\pi (1)}{4.4}\bigg)\\\\B'(t) = 0.09145\\B'(t) \approx 0.09$

The rate of increase after 1 day is 0.09

8 0

2 years ago