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Sav [38]
3 years ago
9

Round 796.301 to the underlined place value.the underlined place value is 0

Mathematics
2 answers:
svetoff [14.1K]3 years ago
6 0
796.300 because the 1 is less than four so it becomes a zero and the numbers in front stay the same.
ollegr [7]3 years ago
3 0
This is a pretty simple all you hav this one's pretty simple all you have to do is Think is zero higher or lower than five or four it is lower than four so if it is lower than four that means it stays the same so you replace all the numbers except the first number with zero
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B : 95! I’m pretty sure cause 90 is a straight angle and that’s near a computers angle is you look at it from the side .
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Find the distance between each pair of points (8,5) ,(-1,3)
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Step-by-step explanation:

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A spring has a natural length of 7 m. If a 4-N force is required to keep it stretched to a length of 11 m, how much work W is re
bezimeni [28]

Answer:

18 J is the work required to stretch a spring from 7 m to 13 m.

Step-by-step explanation:

The work done is defined to be the product of the force F and the distance d  that the object moves:

W=Fd

If F is measured in newtons and d<em> </em>in meters, then the unit for is a newton-meter, which is called a joule (J).

This definition work as long as the force is constant, but if the force is variable like in this case, we have that the work done is given by

W=\int\limits^b_a {f(x)} \, dx

Hooke’s Law states that the force required to maintain a spring stretched x    units beyond its natural length is proportional to

f(x)=kx

where k is a positive constant (called the spring constant).

To find how much work W is required to stretch it from 7 m to 13 m you must:

Step 1: Find the spring constant

We know that the spring has a natural length of 7 m and a 4 N force is required to keep it stretched to a length of 11 m. So, applying Hooke’s Law

4=k(11-7)\\\\\frac{k\left(11-7\right)}{4}=\frac{4}{4}\\\\k=1

Thus F=x

Step 2: Find the the work done in stretching the spring from 7 m to 13 m.

Recall that the natural length is 7 m, so when we stretch the spring from 7 m to 13 m, we are stretching it by 6 m beyond its natural length.

Work needed to stretch it by 6 m beyond its natural length

W=\int\limits^6_0 {x} \, dx \\\\\mathrm{Apply\:the\:Power\:Rule}:\quad \int x^adx=\frac{x^{a+1}}{a+1},\:\quad \:a\ne -1\\\\\left[\frac{x^{1+1}}{1+1}\right]^6_0\\\\\left[\frac{x^2}{2}\right]^6_0=18

18 J is the work required to stretch a spring from 7 m to 13 m.

5 0
3 years ago
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alexira [117]
The answer is the 4th option, 
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7 0
3 years ago
Plllsss answer! I don't understand this question.
Softa [21]

Answer:

  C and D

Step-by-step explanation:

Population density is the ratio of population to area. Its units are persons per square mile. Here, you're being asked to compare the population densities of several countries to the average population density in several US states.

__

<h3>average</h3>

The idea of "average population density of the 10 states listed" is somewhat ambiguous. It could mean either of (a) the ratio of the total of the states' population to the total of their land area, or (b) the average of the population densities of the states. (In the attached, we computed both, but the answer remains the same using either number.)

When there are numerous identical calculations to be performed, it is convenient to let a spreadsheet do them. The attached spreadsheet shows the population densities for the 10 states and 5 countries listed.

Depending on how you define it, the average population density of the 10 states is about 10.5 or about 15.7 people per square mile. (15.7 is the average of the density numbers, found using the spreadsheet AVERAGE function.)

__

<h3>countries</h3>

The 5 countries have population densities ranging from about 7.7 to 236 people per square mile. Two of the countries have density below 10.5, so are the answers to the question asked.

  Canada (C) and Iceland (D) have population density below the US state average.

4 0
2 years ago
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