Answer:
d = 5e
Step-by-step explanation:
Number of wagons = 5
Number of miles per wagon = e
Total number of mules = d
Hence, relationship between total mules d needed will be ;
Total mules = number of wagons * Numbe rof mules per wagon
d = 5e
y < -|x|
replace the letters with the given numbers:
(1,-2) -2<-|1| this is true
(1,-1) -1 <-|1| this is false
(1,0) 0 < -|1| this I false
The answer is (1,-2)
Answer:
Option 1 - 
Step-by-step explanation:
Given : The goals against average (A) for a professional hockey goalie is determined using the formula
. In the formula, g represents the number of goals scored against the goalie and t represents the time played, in minutes.
To find : Which is an equivalent equation solved for g?
Solution :
Solve the formula in terms of g,

Multiply both side by t,

Divide both side by 60,


Therefore, option 1 is correct.
Answer:
(2 x - 3)^2 thus it's True
Step-by-step explanation:
Factor the following:
4 x^2 - 12 x + 9
Factor the quadratic 4 x^2 - 12 x + 9. The coefficient of x^2 is 4 and the constant term is 9. The product of 4 and 9 is 36. The factors of 36 which sum to -12 are -6 and -6. So 4 x^2 - 12 x + 9 = 4 x^2 - 6 x - 6 x + 9 = 2 x (2 x - 3) - 3 (2 x - 3):
2 x (2 x - 3) - 3 (2 x - 3)
Factor 2 x - 3 from 2 x (2 x - 3) - 3 (2 x - 3):
(2 x - 3) (2 x - 3)
(2 x - 3) (2 x - 3) = (2 x - 3)^2:
Answer: (2 x - 3)^2
Answer:
Please check the explanation!
Step-by-step explanation:
Given the polynomial




so expanding summation

solving




also solving






similarly, the result of the remaining terms can be solved such as




so substituting all the solved results in the expression


Therefore,
