So consecutive intergers are 1 away from each other so
the numbers are
x,x+1,x+2
if the product (multily) of 2 smaeller is 5 less than 5 times largest so
(x times (x+1)) is 5 less than 5 times (x+2)
(x times (x+1))=-5+5(x+2)
pemdas
distribute using distributiver property
a(b+c)=ab+ac so
x times (x+1)=x^2+x
5(x+2)=5x+10
we have
x^2+x=-5+5x+10
add like terms
x^2+x=5x+5
subtract (5x+5) from both sides
x^2-4x-5=0
factor by find what 2 numbers add to -4 and multply to get -5
numbers are 1 and -5
so we do
(x+1)(x-5)=0
sete each to zero
x+1=0
x-5=0
x+1=0
sbtract 1
x=-1
we want positie so thisis wrong answer
x-5=0
add 5
x=5
subsitute
x,x+1,x+2
5,5+1,5+2
5,6,7
smalles is 5
Y=mx+b
m=slope
b=yintercept
slope=-3/4
yintercept=3/2
y=-3/4x+3/2
it could also be written
3x+4y=6
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)
