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balandron [24]
4 years ago
8

the perimeter of Jared's property is 56 yards. how many feet of fence will Jared need if he wants to enclose the whole property?

Mathematics
1 answer:
prisoha [69]4 years ago
5 0
There are 3 feet in 1 yard.

Multiply 56 × 3 = 168

Jared will need 168 feet of fence to enclose his property.
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The answer is 26 because I said so I took this 5 years ago I’m in college now
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1. Name three examples of each of the types of properties of matter:
uysha [10]

Intensive properties and extensive properties are types of physical properties of matter. The terms intensive and extensive were first described by physical chemist and physicist Richard C. Tolman in 1917. Here's a look at what intensive and extensive properties are, examples of them, and how to tell them apart.


Intensive Properties

Intensive properties are bulk properties, which means they do not depend on the amount of matter that is present. Examples of intensive properties include:

Boiling point

Density

State of matter

Color

Melting point

Odor

Temperature

Refractive Index

Luster

Hardness

Ductility

Malleability

Intensive properties can be used to help identify a sample because these characteristics do not depend on the amount of sample, nor do they change according to conditions.


Extensive Properties

Extensive properties do depend on the amount of matter that is present. An extensive property is considered additive for subsystems. Examples of extensive properties include:

Volume

Mass

Size

Weight

Length

The ratio between two extensive properties is an intensive property. For example, mass and volume are extensive properties, but their ratio (density) is an intensive property of matter.

While extensive properties are great for describing a sample, they aren't very helpful identifying it because they can change according to sample size or conditions.


Way to Tell Intensive and Extensive Properties Apart

One easy way to tell whether a physical property is intensive or extensive is to take two identical samples of a substance and put them together. If this doubles the property (e.g., twice the mass, twice as long), it's an extensive property. If the property is unchanged by altering the sample size, it's an intensive property.

6 0
3 years ago
Imagine a world in which only decimals, not fractions, are used. How would your life be different?
nataly862011 [7]
I used to hate fractions. But in time, you learn to love them. This is because there's a big difference between fractions and decimals, even though when you divide the actual fraction it comes out to a decimal. Decimals go on and on sometimes, and it would be impossible to write out all those numbers, especially when taking a timed test, for example. Fractions, in this case, would be much more useful (as long as you know how to use them to your advantage). Fractions are basically all those decimal numbers wrapped up into a single, simple division. It makes the outcome of your answer much more accurate than if you estimate every decimal you get throughout a math problem. The more you estimate throughout the problem-solving process, the less accurate your final answer will be. Hence why teachers will usually tell you to estimate when you're putting down the final answer. Fractions are complex at times, so it may be easier to use them in decimal form for certain situations (especially if the decimal form is short and sweet). A world without fractions will result in many, many inaccurate situations involving mathematical knowledge.
3 0
3 years ago
In this problem we consider an equation in differential form Mdx+Ndy=0. (4x+2y)dx+(2x+8y)dy=0 Find My= 2 Nx= 2 If the problem is
zheka24 [161]

Answer:

f(x,y)=2x^2+4y^2+2xy=C_1\\\\Where\\\\y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )

Step-by-step explanation:

Let:

M(x,y)=4x+2y\\\\and\\\\N(x,y)=2x+8y

This is and exact equation, because:

\frac{\partial M(x,y)}{\partial y} =2=\frac{\partial N}{\partial x}

So, define f(x,y) such that:

\frac{\partial f(x,y)}{\partial x} =M(x,y)\\\\and\\\\\frac{\partial f(x,y)}{\partial y} =N(x,y)

The solution will be given by:

f(x,y)=C_1

Where C1 is an arbitrary constant

Integrate \frac{\partial f(x,y)}{\partial x} with respect to x in order to find f(x,y):

f(x,y)=\int\ {4x+2y} \, dx =2x^2+2xy+g(y)

Where g(y) is an arbitrary function of y.

Differentiate f(x,y) with respect to y in order to find g(y):

\frac{\partial f(x,y)}{\partial y} =2x+\frac{d g(y)}{dy}

Substitute into \frac{\partial f(x,y)}{\partial y} =N(x,y)

2x+\frac{dg(y)}{dy} =2x+8y\\\\Solve\hspace{3}for\hspace{3}\frac{dg(y)}{dy}\\\\\frac{dg(y)}{dy}=8y

Integrate \frac{dg(y)}{dy} with respect to y:

g(y)=\int\ {8y} \, dy =4y^2

Substitute g(y) into f(x,y):

f(x,y)=2x^2+4y^2+2xy

The solution is f(x,y)=C1

f(x,y)=2x^2+4y^2+2xy=C_1

Solving y using quadratic formula:

y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )

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kenny6666 [7]
Answers 3 and 4! since there is a negative power, that’s the only time the 0’s are added to the front :)
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