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cestrela7 [59]
3 years ago
9

How do you know if 2x+3y=9xy is linear or nonlinear.

Mathematics
2 answers:
Advocard [28]3 years ago
6 0

We are given equation 2x+3y=9xy.

A linear equation is a equation that has maximum degree of the equation as 1.

Degree is the maximum power(exponent) of the variables.

If two variables are being multiplied together in a term, we can find power of the that term by adding powers of those variables.

In the given equation, on the left side we have two terms 2x and 3y, each opf the variable x, and y has power 1 and right side of the eqaution we have term 9xy. There x and y variables are being multiplied together.

So, the total power of the term would be 1+1=2.

So, the degree of the given equation would be 2.

Because degree of the given equation is not 1. Therefore, given equation is not a linear eqaution and it is a non-linear equation.

Scilla [17]3 years ago
6 0

There are several characteristics that define a linear function:

1) They are of the form y = mx + b, where m and b are real constants, or they can also have the form f (x_{1} , x_{2}, .., x_{n}) = ax_{1} + bx_{2} +, ..., + cx_{n} if the function is of several variables. Where a, b, c are real numbers.

2) The degree of the variable x is always equal to 1 or 0. That is, if there is an expression of the form x^{-1} or x ^ {2}, the function is not linear.

3) Your domain is all real numbers

4) The graph of its function in the xy plane is always a straight line.

Analyzing the aforementioned equation:

The function 2x + 3y = 9xy does not have the form described, since it has a multiplication of two variables (9xy).

The graph of its function in the xy plane is a hyperbola

Your domain is not all real numbers, because the function is not defined for x=\frac{1}{3}

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the area of a rectangular volleyball court is 1800 square feet the courts length is twice it width write a system of equations t
Marianna [84]

Answer:

The dimensions of the rectangular volleyball court are 60 ft x 30 ft

Step-by-step explanation:

Let

x ----> the length of rectangular volleyball court

y ---> the width of the rectangular volleyball court

we know that

The area of the rectangular volleyball court is equal to

A=xy

A=1,800\ ft^2

so

1,800=xy ----> equation A

x=2y -----> equation B

substitute equation B in equation A

1,800=(2y)y

1,800=2y^2

Solve for y

Simplify

900=y^2

take square root both sides

y=30\ ft

<em>Find the value of x</em>

x=2y

substitute the value of y

x=2(30)=60\ ft

therefore

The dimensions of the rectangular volleyball court are 60 ft x 30 ft

7 0
3 years ago
you currently have 24 credit hours and a 2.8 gpa you need a 3.0 gpa to get into the college. if you are taking a 16 credit hours
Juliette [100K]

Answer:

\sum_{i=1}^n w_i *X_i = 2.8*24 = 67.2

And for this case we want a gpa of 3.0 taking in count that in this semester he/ she is going to take 16 credits so then the new mean would be given by:

\bar X_f = \frac{\sum_{i=1}^n w_i *X_i+w_f *X_f }{24+16} = 3.0

And we can solve for \sum_{i=1}^n w_f *X_f and solving we got:

3.0 *(24+16) =\sum_{i=1}^n w_i *X_i +\sum_{i=1}^n w_f *X_f

And from the previous result we got:

3.0 *(24+16) =67.2 +\sum_{i=1}^n w_f *X_f

And solving we got:

\sum_{i=1}^n w_f *X_f =120 -67.2= 52.8

And then we can find the mean with this formula:

\bar X_2 = \frac{\sum_{i=1}^n w_f *X_f}{16}= \frac{52.8}{16}=16=3.3

So then we need a 3.3 on this semester in order to get a cumulate gpa of 3.0

Step-by-step explanation:

For this case we know that the currently mean is 2.8 and is given by:

\bar X = \frac{\sum_{i=1}^n w_i *X_i }{24} = 2.8

Where w_i represent the number of credits and X_i the grade for each subject. From this case we can find the following sum:

\sum_{i=1}^n w_i *X_i = 2.8*24 = 67.2

And for this case we want a gpa of 3.0 taking in count that in this semester he/ she is going to take 16 credits so then the new mean would be given by:

\bar X_f = \frac{\sum_{i=1}^n w_i *X_i+w_f *X_f }{24+16} = 3.0

And we can solve for \sum_{i=1}^n w_f *X_f and solving we got:

3.0 *(24+16) =\sum_{i=1}^n w_i *X_i +\sum_{i=1}^n w_f *X_f

And from the previous result we got:

3.0 *(24+16) =67.2 +\sum_{i=1}^n w_f *X_f

And solving we got:

\sum_{i=1}^n w_f *X_f =120 -67.2= 52.8

And then we can find the mean with this formula:

\bar X_2 = \frac{\sum_{i=1}^n w_f *X_f}{16}= \frac{52.8}{16}=16=3.3

So then we need a 3.3 on this semester in order to get a cumulate gpa of 3.0

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3 years ago
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Who should fill out the W-2 form?
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I think B is the answer. I am not sure
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