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Mrac [35]
3 years ago
12

Find the zeros of polynomial functions and solve polynomial equations. f(x)=x^3-5x^2-9x+45

Mathematics
1 answer:
Novay_Z [31]3 years ago
3 0

Find the zeros, just equal the equation to 0.


x³ - 5x² - 9x + 45 = 0


Factor:


x².(x - 5) - 9.(x - 5) = 0

(x² - 9).(x - 5) = 0


It's a multiplication, so, if one term goes to 0, everything goes too.


x² - 9 = 0

x² = 9

x = + or - √9


x' = 3

x'' = -3


x - 5 = 0

x = 5


Now we have three roots: -3, 3, 5.

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defon
Y = 15

Because it’s going to every time by 15
5 0
3 years ago
BRAINLIEST AWARD NO.2
zzz [600]

Answer:

A whole number first term to render as fifth term a value larger than 10000, should be at least 121

Step-by-step explanation:

The formula is given as recursive since it involves the previous number of the sequence, and defined as:

a_n=a_{n-1}*3+6

we also know that the first term is 4

Then in this case, the first five terms are:

a_1=4\\a_2=4*3+6=18\\a_3=18*3+6=60\\a_4=60*3+6=186\\a_5=186*3+6=564\\

So if we want to find the first term in the case that the fifth one is greater than 10,000 using this recursive formula, now we have to start backwards, and say that the fifth term is "> 10000" and what the fourth one is.

Notice that if you have this definition for the nth term, we can obtain from it, what the previous term is to find the general rule:

a_n=a_{n-1}*3+6\\a_n-6=a_{n-1}*3\\\frac{a_n-6}{3} = a_{n-1}\\a_{n-1}=\frac{a_n}{3} -2

So the rule is to subtract 6 from he term, and divide the subtraction by 3. Then working backwards:

a_5>10000\\\frac{a_5}{3} -2>\frac{10000}{3} -2\\a_4>=\frac{10000}{3} -2\\\frac{a_4}{3} -2>\frac{\frac{10000}{3}-2}{3}-2 =\frac{10000}{9}-\frac{8}{3} \\a_3>\frac{10000}{9}-\frac{8}{3} \\\frac{a_3}{3} -2>\frac{\frac{10000}{9}-\frac{8}{3} }{3} -2=\frac{10000}{27} -\frac{8}{9} -2=\frac{10000}{27} -\frac{26}{9}\\a_2=\frac{10000}{27} -\frac{26}{9}\\\frac{a_2}{3} -2>\frac{\frac{10000}{27} -\frac{26}{9}}{3} -2=\frac{10000}{81} -\frac{80}{27} \\a_1>\frac{10000}{81} -\frac{80}{27}\approx 120.49

therefore, the starting first term should be at least about 121 to give a fifth term larger than 10,000

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Suppose $200 is invested at the annual interest rate 6% compounded continuously what is the amount in the account after years ?
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