So to add fractions they need to have a common denominator, which in this case is 24. then multiply 1 by 4 and 5 by 3 because whatever you multiply the denominator by you need to do the same to the numerator, which will get you 4/24+15/24 then you add across the numerators and you get 19/24
A ↔ B ↔ C ↔ D ↔ E ↔ F
8 7
???
AB + BC + CD = AD <em>segment addition postulate</em>
+ 8 + 7 = AD
+ 15 = AD
AD + 60 = 4AD
60 = 3AD
20 = AD
AB =
=
= 5
DE =
=
= 4
CD + DE + EF = CF <em>segment addition postulate</em>
7 + 4 + EF = CF
11 + EF = CF
Answer: 11 + EF
Note: You did not provide any info about EF. If you have additional information that you did not type in, calculate EF and add it to 11 to find the length of CF.
Answer:
radius x radius x pi = area of circle
6 x 6 = 36
36 x 3.14 = 113.04
113.04
Hope this helps
Step-by-step explanation:
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}](https://tex.z-dn.net/?f=x1%2Cx2%3D%5Cfrac%7B-b%2B-%5Csqrt%7Bb%5E%7B2%7D-4ac%20%7D%20%7D%7B2a%7D)
<h3>This case</h3>
The quadratic function is f(x) = x² + 4x +9
Being:
the zeros or roots are calculated as:
![x1=\frac{-4+\sqrt{4^{2}-4x1x9 } }{2x1}](https://tex.z-dn.net/?f=x1%3D%5Cfrac%7B-4%2B%5Csqrt%7B4%5E%7B2%7D-4x1x9%20%7D%20%7D%7B2x1%7D)
![x1=\frac{-4+\sqrt{16-36 } }{2x1}](https://tex.z-dn.net/?f=x1%3D%5Cfrac%7B-4%2B%5Csqrt%7B16-36%20%7D%20%7D%7B2x1%7D)
![x1=\frac{-4+\sqrt{-20 } }{2x1}](https://tex.z-dn.net/?f=x1%3D%5Cfrac%7B-4%2B%5Csqrt%7B-20%20%7D%20%7D%7B2x1%7D)
and
![x2=\frac{-4-\sqrt{4^{2}-4x1x9 } }{2x1}](https://tex.z-dn.net/?f=x2%3D%5Cfrac%7B-4-%5Csqrt%7B4%5E%7B2%7D-4x1x9%20%7D%20%7D%7B2x1%7D)
![x2=\frac{-4-\sqrt{16-36 } }{2x1}](https://tex.z-dn.net/?f=x2%3D%5Cfrac%7B-4-%5Csqrt%7B16-36%20%7D%20%7D%7B2x1%7D)
![x2=\frac{-4-\sqrt{-20} }{2x1}](https://tex.z-dn.net/?f=x2%3D%5Cfrac%7B-4-%5Csqrt%7B-20%7D%20%7D%7B2x1%7D)
If the content of the root is negative, the root will have no solution within the set of real numbers. Then
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
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