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bixtya [17]
2 years ago
15

Help please

Mathematics
1 answer:
Setler79 [48]2 years ago
4 0

Answer: inconsistent

Step-by-step explanation:

Inconsistent systems have two lines that are two independent equations that don’t overlap. Basically, if the lines are parallel, the system will always be inconsistent.

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Help me please will mark brainlyist
Leni [432]

Answer:

24.46

Step-by-step explanation:

4 0
3 years ago
Find f^-1(x) for f(x) = 1/x^3
kow [346]

Answer: f^{-1}(x) = \frac{\sqrt[3]{x^{2}}}{x}

<u>Step-by-step explanation:</u>

y = \frac{1}{x^{3}}

Inverse is when you swap the x's and y's and then solve for "y":

x = \frac{1}{y^{3}}

y^{3} = \frac{1}{x}

y = \frac{1}{\sqrt[3]{x}}

y = \frac{1}{\sqrt[3]{x}}*(\frac{\sqrt[3]{x}}{\sqrt[3]{x}})^{2}

y = \frac{\sqrt[3]{x^{2}}}{x}

4 0
3 years ago
A rectangle has a length that is 3 units greater than its width. What algebraic expression describes its perimeter
harina [27]
This expression should do

L=3+W
7 0
3 years ago
What is the solution to 4/9*-10&gt;*/3-12?
luda_lava [24]

The answer is 2 ( choose 2 )

and how we find this answer? the picture is clearly for that.

8 0
3 years ago
What are the roots of this equation? x2-4x+9=0
zysi [14]

Considering the definition of zeros of a function, the zeros of the quadratic function f(x) = x² + 4x +9 do not exist.

<h3>Zeros of a function</h3>

The points where a polynomial function crosses the axis of the independent term (x) represent the so-called zeros of the function.

In summary, the roots or zeros of the quadratic function are those values ​​of x for which the expression is equal to 0. Graphically, the roots correspond to the abscissa of the points where the parabola intersects the x-axis.

In a quadratic function that has the form:

f(x)= ax² + bx + c

the zeros or roots are calculated by:

x1,x2=\frac{-b+-\sqrt{b^{2}-4ac } }{2a}

<h3>This case</h3>

The quadratic function is f(x) = x² + 4x +9

Being:

  • a= 1
  • b= 4
  • c= 9

the zeros or roots are calculated as:

x1=\frac{-4+\sqrt{4^{2}-4x1x9 } }{2x1}

x1=\frac{-4+\sqrt{16-36 } }{2x1}

x1=\frac{-4+\sqrt{-20 } }{2x1}

and

x2=\frac{-4-\sqrt{4^{2}-4x1x9 } }{2x1}

x2=\frac{-4-\sqrt{16-36 } }{2x1}

x2=\frac{-4-\sqrt{-20} }{2x1}

If the content of the root is negative, the root will have no solution within the set of real numbers. Then \sqrt{-20} has no solution.

Finally, the zeros of the quadratic function f(x) = x² + 4x +9 do not exist.

Learn more about the zeros of a quadratic function:

brainly.com/question/842305

brainly.com/question/14477557

#SPJ1

6 0
1 year ago
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