Ask question

Ask question

All categories

2 years ago

8

find the root(s) of f (x) = (x + 5)3(x - 9)2(x + 1). 50 POINTS!!!!!!! -5 with multiplicity 3 5 with multiplicity 3 -9 with multi

plicity 2 9 with multiplicity 2 -1 with multiplicity 0 -1 with multiplicity 1 1 with multiplicity 0 1 with multiplicity 1

2 answers:

Dmitrij [34]2 years ago

8 0

Answer:

The answer to this question can be described as follows:

-5 with multiplicity 3

9 with multiplicity 2

-1 with multiplicity 1

Step-by-step explanation:

Given:

$\Rightarrow f (x) = (x + 5)^3(x - 9)^2(x + 1)$

Solve the above equation:

$\Rightarrow f (x) = (x + 5)(x + 5)(x + 5)(x - 9)(x - 9)(x + 1)\\$

The roots of the polynomial are as follows:

$\Rightarrow (x-x_1)(x-x_2)(x-x_3)......(x-x_n)\\\\\Rightarrow x_1,x_2,x_3.........x_n$

That's why the roots are:

5 with multiplicity 3

9 with multiplicity 2

-1 with multiplicity 1

AysviL [449]2 years ago

5 0

Answer:

answer:

a. -5 with multiplicity 3

d. 9 with multiplicity 2

f. -1 with multiplicity 1

Step-by-step explanation:

You might be interested in

Ted is a long distance driver. It took him 30 min longer to drive 275km on the Trans-Canada highway west of Swift Current, Saska

IRISSAK [1]

Speed traveling west = x - 10 km /hr

275 / ( x-10) - 300 / x = 1/2

275 *2x - 300 (x-10)*2 = ( x-10) *x

550x - 600x +6000 = x^2 -10x

x^2 +40x - 6000 = 0

( x -60) ( x+100) = 0

x = 60

ANSWER East = 60 km/ hr West = 50 km /hr

CHECK

300 /60 = 5 hours

275 /50 = 5.5 hours

7 0

2 years ago

Use a table of numerical values of f(x,y) for (x,y) near the origin to make a conjecture about the value of the limit of f(x,y)

grigory [225]

Seems to be that the limit to compute is

$\displaystyle\lim_{(x,y)\to(0,0)}\frac{xy}{x^2+2y^2}$

Consider an arbitrary line through the origin $y=mx$ , so that we rewrite the above as

$\displaystyle\lim_{x\to0}\frac{mx^2}{x^2+2m^2x^2}=\lim_{x\to0}\frac m{1+2m^2}=\frac m{1+2m^2}$

The value of the limit then depends on the slope $m$ of the line chosen, which means the limit is path-dependent and thus does not exist.

8 0

2 years ago

Read 2 more answers

A health food company sells packs of assorted nuts. The company incurs a monthly fixed cost of $1,400 for ingredients and raw ma

tino4ka555 [31]

If you're asking about what the equation would be written as, then it's y = 1x + 1,400.

8 0

2 years ago

Read 2 more answers

Billy graphed the system of linear equations to find an approximate solution. y = A system of equations. y equals negative Start

Elina [12.6K]

1.09!!!!!!!!!!!!!!!!!!!

3 1

2 years ago

Read 2 more answers

In 154 N38 what digit should be replaced to N to make the number divisible by 3,6 and 9?

JulsSmile [24]

Answer:

$N = 6$ so that number becomes divisible by 3, 6 and 9.

Step-by-step explanation:

In Number Theory there is a rule of thumb which states that sum of digits of a multiple of 3 equal 3 or a multiple of three. If we know that $n = 154N38$ , then its sum of digits is:

$x = 1 + 5+4+N+3+8$

$x = 21+N$ (Eq. 1)

We have to determine which digits corresponds to multiples of three, there are four digits:

N = 0

$x = 21+0$

$x = 21$ ( $3\times 7 = 21$ )

N = 3

$x = 21+3$

$x = 24$ ( $3\times 8 = 24$ )

N = 6

$x = 21+6$

$x = 27$ ( $3\times 9 = 27$ )

N = 9

$x = 21+9$

$x = 30$ ( $3\times 10 = 30$ )

We get the following four distinct options: 154038, 154338, 154638, 154938. Now we find which number is divisible by 6 and 9 by factor decomposition:

$154038 = 2\times 3\times 25673$

$154338 =2\times 3\times 29 \times 887$

$154638 = 2\times 3\times 3\times 11\times 11\times 71$

$154938 = 2\times 3\times 7\times 7\times 17\times 31$

It is quite evident that $N = 6$ so that number becomes divisible by 3, 6 and 9.

6 0

2 years ago