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3 years ago

A neutron star has a density of about 5.9 10 17 kg/m3.

2 answers:

7 0

Approximate mass = 5.9*10^11 kg

Density = Mass/Volume
5.9*10^17 kg/m^3 = (unknown=Mass)/1 cm^3 (this must be converted first into m^3 to have unit consistency)
1 cm^3 (1/100)^3 = 1*10^-6 m^3

(5.9 * 10 ^17 kg/m^3)(1*10^-6 m^3) = Mass
5.9 * 10 ^ 11 kg = Mass

Andru [333]3 years ago

6 0

The answer is, 5.9 10 ^11 kg (1.3 trillion pounds)

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In which quadrant will the image FGH lie after a counterclockwise rotation of 1980 degrees? Explain how you made your prediction

Nata [24]

A 360 degrees counter-clockwise rotation will bring the triangle at the same place.

So, divide 1980 by 360 and find the remainder.

On dividing, we can find that

1980 = 5(360) + 180

A 1800 degrees rotation will rotate the triangle 5 times and will bring it in the same place.

A further 180 degrees rotation will rotate the triangle to two quadrants back and will bring it in the first quadrant.

3 0

3 years ago

The ratio of the number of apples to the number of pears in a basket was 3:5. After 7more apples were added to the basket and 9

atroni [7]

Answer:

Apples = 24

Pears = 40

Step-by-step explanation:

Let u be the unit of fruit in the basket,

number of apples: $3u +7$

number of pears: $5u -9$

now to find u, we equate the two equations.

$3u+7=5u-9$

Now simplify.

$5u-3u=7+9$

$2u=16$

$u=8$

Now we can find the original number of apples and pears by substituting u.

so original number of apples = $3u = 3(8) = 24$

and original number of pears = $5u = 5(8) = 40$

5 0

4 years ago

I’m which place should you put the first digit in the quotient in division

Sergio [31]

Step-by-step explanation:

It would be the very top of the division problem, so for example, if we take. a look at the lower image below, we see that the number 6 is the number that would be the (first) number that would be the quotient.

Answer:

Very top, (e.g)<em> "the number 6"</em>

6 0

3 years ago

The frequency table below shows the antenna careers among the incoming class of first-year college students. Every student is ch

neonofarm [45]

To solve this you must use a proportion like so...

$\frac{part}{whole} = \frac{part}{whole}$

The total number of students that can be chosen are 4,663. This number will represent the whole of one fraction in the proportion. We want to know what percent probability out of these students are engineer, medical doctor/surgeon. This would be considered the part of this fraction. Sum the number of engineering students (615) with medical doctors/surgeons (723) to find this number

723 + 615 = 1,338 students that want to be an engineer or medical doctor/surgeon

Percent's are always taken out of the 100. This means that the other fraction in the proportion will have 100 as the whole and x (the unknown) as the part.

Here is your proportion:

$\frac{1,338}{4,663} =\frac{x}{100}$

Now you must cross multiply

1,338*100 = 4,663*x

133,800 = 4,663x

To isolate x divide 4,663 to both sides

133,800/4,663 = 4,663x/4,663

28.7 = x

This means that there is a 28.7% of a student with the intent of becoming an engineer or a medical doctor/surgeon to be chosen at random

Hope this helped!

~Just a girl in love with Shawn Mendes

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3 years ago

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Maksim231197 [3]

Answer:

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Step-by-step explanation:

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3 years ago