Consecutive integers are 1 apart
x,x+1,x+2
(x)(x+1)(x+2)=-120
x^3+3x^2+3x=-120
add 120 to both sides
x^3+3x^2+3x+120=0
factor
(x+6)(x^2-3x+20)=0
set each to zero
x+6=0
x=-6
x^2-3x+20=0
will yeild non-real result, discard
x=-6
x+1=-5
x+2=-4
the numbers are -4,-5,-6
use trial and error and logic
factor 120
120=2*2*2*3*5
how can we rearange these numbers in (x)(y)(z) format such that they multiply to 120?
obviously, the 5 has to stay since 2*5=10 which is out of range
so 2*2*2*3 has to arrange to get 3,4 or 4, 6 or 6,7
obviously, 7 cannot happen since it is prime
3 and 4 results in in 12, but 2*2*2*3=24
therfor answer is 4 and 6
they are all negative since negaive cancel except 1
the numbers are -4,-5,-6
You use the distance distance formula d √︎(x₂-x₁)+(y₂-y₁) on this as well as y=mx+b.
With mx+b you find your second set of coordinates. With the distance formula you get your square root.
So y=-3x+10 well send you 10 up the y intercept the, neg 3 down, and positive one across the x axis.
This gives you your second set of coordinates (1,7) to go with your first set.
You apply both sets of coordinates to the distance formula. d √︎(x₂-x₁)+(y₂-y₁)
You get √︎(1-(-6))+(7-8)
Then √︎7-1
Finally √︎6
And √︎6 is between 2*2 and 3*3
Rough guess, the Sq root of 6 is 2.449 or rounded to the hundredth 2.45
Answer:
69.36
Step-by-step explanation:
Multiply 289 with 24 and divide the result by 100
289 × 24 ÷ 100 = 69.36
