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Delvig [45]
2 years ago
5

This is nothing you need to solve, I just need a refresher about this type of math. For some reason I can't remember this right

now. Can you please help?
So if I have 2x^2 + 1x, isn't there technically an exponent of 1 above the 1x?
(I'm trying to factor and solve the equation: 2x^2 + x - 21 = 0, if that helps.)
Mathematics
2 answers:
GrogVix [38]2 years ago
8 0

Answer and Step-by-step explanation:

As goddesboi mentioned, x^1 is the same as x.

Because of this, we don't put the 1 in the exponents place, but if you do, there is nothing wrong with it. This can also be applied to 1 times a number, say for example <em>x</em>. We don't keep the 1 with the x because it is the same thing as saying, x, so it is easier and faster to write.

<u><em>I hope this helps!
#teamtrees #PAW (Plant And Water)
</em></u>

SVEN [57.7K]2 years ago
6 0

Answer:

Yes there is (but we like to remove it)

Step-by-step explanation:

Recall that x^1 = x, which represents x being multiplied by itself once, so you just have x. So, we usually take out the exponent 1 because it is not necessary since it's just the same as x.

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The graph of y=x^2-4x is shown. name a zero of the function below
hoa [83]

Answer:

0 and 4

Step-by-step explanation:

y = x^2 - 4x can be factored as follows:  y = x(x - 4).

We find the zeros by setting this factored form equal to zero and solving for x:

x(x - 4) = 0, or x = 0, or x = 4.

The two zeros of this function are 0 and 4.

5 0
3 years ago
The (i,j) minor of a matrix A is the matrix Aij obtained by deleting row i and column j from A.
STatiana [176]

The (i,j) minor of a matrix A is the matrix Aij obtained by deleting row i and column j from A it is true.

The (i,j) minor of a matrix A is the matrix Aij obtained by deleting row i and column j from A. A determinant of an n×n matrix can be defined as a sum of multiples of determinants of (n−1)×(n−1) sub matrices.

This is done by deleting the row and column which the elements belong and then finding the determinant by considering the remaining elements. Then find the co factor of the elements. It is done by multiplying the minor of the element with -1i+j. If Mij is the minor, then co factor, c_{ij} = -1^{i+j} + m_{ij}.

Each element in a square matrix has its own minor. The minor is the value of the determinant of the matrix that results from crossing out the row and column of the element .

Learn more about the minor of the matrix here:

brainly.com/question/26231126

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4 0
1 year ago
Solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value y
zalisa [80]

Answer:

y has a finite solution for any value y_0 ≠ 0.

Step-by-step explanation:

Given the differential equation

y' + y³ = 0

We can rewrite this as

dy/dx + y³ = 0

Multiplying through by dx

dy + y³dx = 0

Divide through by y³, we have

dy/y³ + dx = 0

dy/y³ = -dx

Integrating both sides

-1/(2y²) = - x + c

Multiplying through by -1, we have

1/(2y²) = x + C (Where C = -c)

Applying the initial condition y(0) = y_0, put x = 0, and y = y_0

1/(2y_0²) = 0 + C

C = 1/(2y_0²)

So

1/(2y²) = x + 1/(2y_0²)

2y² = 1/[x + 1/(2y_0²)]

y² = 1/[2x + 1/(y_0²)]

y = 1/[2x + 1/(y_0²)]½

This is the required solution to the initial value problem.

The interval of the solution depends on the value of y_0. There are infinitely many solutions for y_0 assumes a real number.

For y_0 = 0, the solution has an expression 1/0, which makes the solution infinite.

With this, y has a finite solution for any value y_0 ≠ 0.

8 0
4 years ago
What is the degree of the polynomial f(x)=5x^4 -7x^3-12x^2+4x-2?
allochka39001 [22]
The degree of the polynomial is 4.

Explanation: Look at the highest degree in the polynomial. The highest degree in the given polynomial is 4. (The 5x^4 contains the highest degree which is, of course, 4.)
I hope this helps.
5 0
3 years ago
What 3 to the 6th power
abruzzese [7]

Answer:

729

Step-by-step explanation:

8 0
3 years ago
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