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Aloiza [94]
3 years ago
5

The formula for the area of a square is s2, where s is the side length of the square. What is the area of a square with a side l

ength of 6 centimeters? Do not include units in your answer.
Mathematics
1 answer:
nignag [31]3 years ago
5 0

Answer:

36

Step-by-step explanation:

If s is the side length, then the area of the square is

A = s^ = (6)^2 = 36

This is how you were expected to answer this question.  However, the actual correct answer is 36 cm^2.

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How to graph -x + 2y = 6
larisa86 [58]
In order to graph the equation, you need some coordinates, 
So, put x = 0, 
2y = 6. y = 3.  Coordinate = (0, 3)

Now, put y = 0, 
-x = 6
x = -6   Coordinate (-6, 0)

In short, Just mark that two points and draw a line, graph is done

Hope this helps!
3 0
3 years ago
What is the difference, in hours and minutes, between these times?
elena-s [515]

Answer:

jhj

Step-by-step explanation:

4 0
3 years ago
What’s 45 in scientific notation
Mashcka [7]

Answer

is 45 written as 4.5 x 101 in scientific notation

Step-by-step explanation:

Can i Be mark as the brainliest answer PLEASE!

8 0
3 years ago
Read 2 more answers
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
I need help on 5 - 11
nalin [4]

5-11.

Positive 5 minus negative 11=

11-5=6

Now is 11 greater or 5, 11 is. And what is the integer of 11? A negative.

Therefore, the answer is -6.

4 0
2 years ago
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