Answer:
AB+BE+EF+GF+HG+AH = 14 #EQU 1
BC+CD+DE+EF+GF+BG = 22 #EQU 2
AB+BC+CD+DE+EF+GF+HG+AH = 24 #EQU 3
IMPLEMENTING EQU 2 IN EQU 3
AB+22-EF+ HG+AH = 24 # EQU 5
FROM EQU 1
AB+HG+AH = 14-(BE+EF+GF) #EQU 4
IMPLEMENTING EQU 4 IN EQU 5
22+14 - (BE+EF+GF+EF) = 24
36 - 24 = BE+EF+GF+BG(EF=BG)
12 = PATH TAKEN ON THURSDAY
Answer:5x - 6
Step-by-step explanation:
We know for our problem that the zeroes of our quadratic equation are

and

, which means that the solutions for our equation are

and

. We are going to use those solutions to express our quadratic equation in the form

; to do that we will use the <span>zero factor property in reverse:
</span>



<span>
</span>



<span>
Now, we can multiply the left sides of our equations:
</span>

<span>= </span>

=

=

Now that we have our quadratic in the form

, we can infer that

and

; therefore, we can conclude that

.
Answer:
24
Step-by-step explanation:
multiply 10 by 1.5 to get 15.
do the same with 16
Answer:
Therefore the Final Solution is
x = 1 and y = 1
Step-by-step explanation:
Given:
...............Equation ( 1 )
................Equation ( 2 )
To Find:
x = ?
y = ?
Solution:
Equating Equation ( 1 ) in Equation ( 2 ) we get

Now Substituting y = 1 in Equation ( 1 ) we get

Therefore the Final Solution is
x = 1 and y = 1