Answer:
84 is the highest possible course average
Step-by-step explanation:
Total number of examinations = 5
Average = sum of scores in each examination/total number of examinations
Let the score for the last examination be x.
Average = (66+78+94+83+x)/5 = y
5y = 321+x
x = 5y -321
If y = 6, x = 5×6 -321 =-291.the student cannot score -291
If y = 80, x = 5×80 -321 =79.he can still score higher
If If y = 84, x = 5×84 -321 =99.This would be the highest possible course average after the last examination.
If y= 100
The average cannot be 100 as student cannot score 179(maximum score is 100)
Answer: B I believe
Step-by-step explanation:
Answer:
x = 6, x = - 6
Step-by-step explanation:
Given
y = x² - 36
To find the zeros let y = 0, that is
x² - 36 = 0 ← x² - 36 is a difference of squares and factors in general as
a² - b² = (a - b)(a + b), thus
x² - 36 = 0
x² - 6² = 0
(x - 6)(x + 6) = 0 ← in factored form
Equate each factor to zero and solve for x
x - 6 = 0 ⇒ x = 6
x + 6 = 0 ⇒ x = - 6
Step-by-step explanation:
AC+CB+BA=180
2x+23+3x-8+4x+1=180
9x+16=180
9x=180-16
9x=164
x=164/9
x=18.2
Ok well what I did but I’m not sure if it answers your question but I did 27-15=12-7=5 and that what I got maybe it’s 5 it’s just an assumption
