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aniked [119]
3 years ago
12

I do not understand how to do this?

Mathematics
2 answers:
djverab [1.8K]3 years ago
4 0

Answer:

try pressing the help button, it looks like it mightve glitched

Step-by-step explanation:

laila [671]3 years ago
3 0

Answer:

-8,-7

Step-by-step explanation:

I think

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What is sum of the vectors, (4.-5) and (10,2)?<br><br> (-6, 7)<br> (14,-3)<br> (14,7)<br> (15,6)
balu736 [363]

Answer:

(14,-3)

Step-by-step explanation:

To sum vectors, simply sum values at each position.

The sum is (4+10, -5+2) = (14,-3)

8 0
2 years ago
Read 2 more answers
How are the real solutions of a quadratic equation related to the graph of the quadratic function?
Anika [276]
The solutions you get when you solve the formula are the corresponding y coordinates to your x value. So say a point on your graph is (2,3). The first number is x and the second is y. (x,y). The number you plug into your function is x,or in this case: 2. The solution to the equation when the x value is plugged in is y, or 3. Therefore, giving you a point on your graph.
7 0
3 years ago
#14
Delicious77 [7]

Answer:

The answer is equivalent

5 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
Solve:<br><br> r<br> 5<br> - 6 = -1
Leona [35]
That's right, 5- 6= -1
3 0
3 years ago
Read 2 more answers
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