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yanalaym [24]
4 years ago
13

You look over the songs in a jukebox and determine that you like 17 of the 54 songs.

Mathematics
1 answer:
kkurt [141]4 years ago
7 0
An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or …show more content… An observation of the Moon was conducted from Friday, November 8, 2013 to Thursday, November 14, 2013. The study of the Moon during this period occurred consistently between the hours of 8 and 9 p.m. EST within the Northern Hemisphere at 37.3346° N, 79.5228° W (Bedford, V.A.). The Moon was noted to be illuminated on the right side and had a dark shadow on the left side indicating a waxing phase. The light region grew over the surface of the Moon with each subsequent night. The first night’s phase was waxing crescent with over 25 percent of the Moon lit up. The next night, the light had grown to cover more of the Moon as it continued through its waxing crescent phase. On November 10th, the Moon exhibited traits of being at first-quarter or half-moon status because at least 50 percent of its surface was illuminated. In the following nights, the Moon displayed characteristics of waxing gibbous as the light continued to grow across the moon’s surface from right to left. The Moon was nearing closer to the full moon phase on November 14th as only a very small dark shadow was visible on the left side. The Moon takes 27.3 days (sidereal month) to complete its actual orbit around the Earth. Like the Sun, the Moon rises in the east and sets in the west each day. 
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|-2x + 4| > or equal to 4
gulaghasi [49]

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7 0
4 years ago
HEY! GOT A BUTT LOAD OF HW DUE TOMORROW PLS HELP BRAINLIEST ANSWER INCLUDED! Solve. How much pure acid is in 520 milliliters of
Rama09 [41]

540ml  with 12% solution

get 12% of 540

540 times 0.12 = 64.8 ml of pure acid

3 0
4 years ago
Help please i dont get it ( i dont pay attention in class...)
ss7ja [257]

Answer:

Okay, I won't give the answers... HOWEVER, I will show you how to get them yourself! :>

Step-by-step explanation:

A triangles area is essentially half of a square. In my math class, we would multiply the main base, and the height and then proceed to break it in half. For example;

TRIANGLE ONE-- 16 x 16 =__ then cut your answer in half

<em>Hope I helped you out! ~Chaos</em>

6 0
3 years ago
Read 2 more answers
Estimate this product. 89.65 x 4.75 A 4,500 B 450 C 45 D 4.5 plz hurry
Alona [7]
<u>The correct answer</u> is B.

Explanation:
89.65 is nearly 90.
4.75 is nearly 5.

To estimate the product, we will multiply 90*5, which equals 450.

7 0
3 years ago
Read 2 more answers
AP CALC HELP!!! Worth 38 points. AP Calc AB differental FRQ
sveta [45]

Answer:

5a.  approximately 6 grams remain after 2 seconds

5b.  The graph shown cannot be a solution. The solution has negative slope everywhere.

5c.  y = 50/(t+5)

5d.  The amount is changing at a decreasing rate. (As y gets smaller, so does the magnitude of dy/dt.)

Step-by-step explanation:

5a. The tangent line has the equation ...

  y = f'(0)t +f(0)

Here, that is

  y = -0.02·10²·t +10 = 10 -2t

Then at t=2, the value is ...

  y = 10 -2·2 = 6 . . . . grams remaining

__

5b. y² is always positive (or zero), so -0.02y² will be negative. This is dy/dt, the slope of the curve with respect to time, so any curve with positive slope somewhere cannot be a solution.

__

5c. The equation is separable so can be solved by integrating ...

  ∫y^-2·dy = -0.02∫dt

  -y^-1 = -0.02t +c . . . . for some arbitrary constant c

Multiplying by -50 gives ...

  50/y = t + c . . . . for some constant c

We can find the value of c by invoking the initial condition. At t=0, y=10, so we have ...

  50/10 = 0 +c = 5

Then, solving for y, we get ...

  y = 50/(t+5)

__

5d. As noted above (and as described by the differential equation), the magnitude of the rate of change is proportional to the square of y. As y decreases, its rate of change will also decrease (faster). You can see that the curve for y flattens out as t increases. The amount of the substance is changing at a decreasing rate.

5 0
4 years ago
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