Answer:
Distance of JK = 15 unit
Step-by-step explanation:
Given:
J(4,8)
K(-1,-2)
Find:
Distance of JK
Computation:
Distance = √(x₂-x₁)² + (y₂-y₁)²
Distance of JK = √(-1-4)² + (-2-8)²
Distance of JK = √25 + 100
Distance of JK = √125
Distance of JK = 15 unit
First one is
(2x+5)
Second one is
“All of these are correct”
Hope this helped
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.
The answer is 14 divided by 2