Not understanding what to do need gelp
2 answers:
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Answer:
1) (x + 3)(3x + 2)
2) x= +/-root6 - 1 by 5
Step-by-step explanation:
3x^2 + 11x + 6 = 0 (mid-term break)
using mid-term break
3x^2 + 9x + 2x + 6 = 0
factor out 3x from first pair and +2 from the second pair
3x(x + 3) + 2(x + 3)
factor out x+3
(x + 3)(3x + 2)
5x^2 + 2x = 1 (completing squares)
rearrange the equation
5x^2 + 2x - 1 = 0
divide both sides by 5 to cancel out the 5 of first term
5x^2/5 + 2x/5 - 1/5 = 0/5
x^2 + 2x/5 - 1/5 = 0
rearranging the equation to gain a+b=c form
x^2 + 2x/5 = 1/5
adding (1/5)^2 on both sides
x^2 + 2x/5 + (1/5)^2 = 1/5 + (1/5)^2
(x + 1/5)^2 = 1/5 + 1/25
(x + 1/5)^2 = 5 + 1 by 25
(x + 1/5)^2 = 6/25
taking square root on both sides
root(x + 1/5)^2 = +/- root(6/25)
x + 1/5 = +/- root6 /5
shifting 1/5 on the other side
x = +/- root6 /5 - 1/5
x = +/- root6 - 1 by 5
x = + root6 - 1 by 5 or x= - root6 - 1 by 5
We have that
<span>p(t)=-3t^2+18t-4
using a graphing tool, we can see the maximum of the graph
(see the attached figure)
A) </span><span>In what year of operation does Mr. Cash’s business show maximum profit?
</span>Mr. Cash’s business show maximum profit at year 3 (maximum in the parabole)
<span>B) What is the maximum profit?
23 (hundred of thousand of dollars) = 2.300.000 dollars
</span>c) What time will it be two late?
(This is the time when the graph crosses zero and the profits turn into losses )
5.77 years, or an estimate of about 69 months.
<span>p(t)=-3t^2+18t-4
using a graphing tool, we can see the maximum of the graph
(see the attached figure)
A) </span><span>In what year of operation does Mr. Cash’s business show maximum profit?
</span>Mr. Cash’s business show maximum profit at year 3 (maximum in the parabole)
<span>B) What is the maximum profit?
23 (hundred of thousand of dollars) = 2.300.000 dollars
</span>c) What time will it be two late?
(This is the time when the graph crosses zero and the profits turn into losses )
5.77 years, or an estimate of about 69 months.