Answer: The correct statements are
The GCF of the coefficients is correct.
The variable c is not common to all terms, so a power of c should not have been factored out.
David applied the distributive property.
Step-by-step explanation:
GCF = Greatest common factor
1) GCF of coefficients : (80,32,48)
80 = 2 × 2 × 2 × 2 × 5
32 = 2 × 2 × 2 × 2 × 2
48 = 2 × 2 × 2 × 2 × 3
GCF of coefficients : (80,32,48) is 16.
2) GCF of variables :(
)
= b × b × b × b
= b × b
=b × b × b × b
GCF of variables :() is
3) GCF of 
and c: c is not the GCF of the polynomial. The variable c is not common to all terms, so a power of c should not have been factored out.
4) 
David applied the distributive property.
Answer:
Option (B). Perimeter of the quadrilateral ABCD= 14.6 units
Step-by-step explanation:
From the figure attached,
Coordinates of the vertices are A(3, 5), B(1, 3), C(3, -1), D(5, 3).
Length of AB = 
= 
= 
= 2.83 units
Length of AD = 
=
= 2.83 units
Length of BC = 
= 
= 4.47 units
Length of DC = 
=
= 4.47 units
Perimeter of the quadrilateral = AB + AD + DC + BC
= 2.83 + 2.83 + 4.47 + 4.47
= 14.6 units
Option (B) is the answer.
Answer:
<em>Look Below In the Explanation</em>
Step-by-step explanation:
Point P has coordinates of (4,4) and point Q has coordinates of (4,-4).
Since both points have the same x-coordinates we can subtract point Q's y-coordinate from point P's y-coordinate.
I will set the equation up and try and figure this part by yourself. I hope you learned what to do from previous questions that are very similar to this that I answered.
4 - ( -4) = length of the bed of flowers
Hope that helps and maybe earns a brainliest!
Have a splendid day! :^)
Step-by-step explanation:
General form of a conic section is:
Ax² + Bxy + Cy² + Dx + Ey + F = 0
If B = 0:
Parabola: Either x² or y² term, but not both
Circle: x² and y² have the same coefficient
Ellipse: x² and y² have different positive coefficients
Hyperbola: x² and y² have different signs
Otherwise, look at the discriminant.
If B² − 4AC < 0, then the conic is an ellipse.
If B² − 4AC = 0, then the conic is a parabola.
If B² − 4AC > 0, then the conic is a hyperbola.
