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3 years ago

According to the Fundamental Theorem of Algebra, how many roots exist for the polynomial function?

1 answer:

4 0

It should have 3 solutions.
The number of solutions depends on the term with the highest exponent.
So, in this case we have a cubic equation, there are 3 solutions.
If we have a quartic equation (4th power), there should be 4 solutions.

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From 93 books to 128 books The percent increase is %

sasho [114]

Im pretty sure that you just take 128 devided by 98 and more the decimal to get the result of 13%

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4 years ago

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Any mind helping? 15 Points! :>

Hitman42 [59]

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3 years ago

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4. "Ten less than the quotient of a number and 3 is 8."

Nookie1986 [14]

Answer:

3x - 10 = 8

Step-by-step explanation:

Let's reconsider the wording of this problem.

"Ten less than the quotient of a number and

3 is 8."

From less, we can assume 10 is being subtracted from something.

We have:

-10

Next, we know that the quotient is the sum of a multiplication problem. So, we know 3 has to be multiplied by something. We can see that the text says 'a number'. If they don't give us the value we can guess that it is the variable x.

Now:

3x - 10

From 'is 8' we know that must be the final sum of the equation.

Finally:

3x - 10 = 8

7 0

3 years ago

Hi please i nedd help with these questions .

Vlad1618 [11]

Answer:

Step-by-step explanation:

#1:

$\frac{18m^2u}{16n^3v^2} \div \frac{24m}{15nu^3}*\frac{8n^2v^3}{30m^3u}$

division sign means that we flip the fraction

$\frac{18m^2u}{16n^3v^2} * \frac{15nu^3}{24m}*\frac{8n^2v^3}{30m^3u}$

now we can multiply all the constants together and all variables

$m: (m^2)/(m*m^4) = (m^2)/(m^4) = 1/(m^2)\\u: (u * u^3)/(u) = (u^4)/(u) = u^3\\n: (n*n^2)/(n^3) = 1\\v: (v^3) / (v^2) = v\\(18*15*8)/(16*24*30) = \frac{3}{16}$

now we can combine all the parts

$\frac{3u^3v}{16m^2}$

#2:

$\frac{24a^2b^3c}{9bc^3}\div\frac{4a^5bc^3}{27a^3b^2c}$

$\frac{24a^2b^3c}{9bc^3}*\frac{27a^3b^2c}{4a^5bc^3}$

$a: (a^2 * a^3)/(a^5) = 1\\b: (b^3 *b^2)/(b*b) = b^3\\c: (c*c)/(c^3*c^3)= 1/(c^4)\\(24*27)/(9*4)= 18$

$\frac{18b^3}{c^4}$

3 0

3 years ago

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Find the coordinates of the midpoint of a segment with the given endpoints.

Step2247 [10]

Answer:

a. (1,2), b. (1, 0)

Step-by-step explanation:

$\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\\\text{Substitute the values.}\\\frac{-3+5}{2},\frac{5+(-1)}{2}\\\text{Evaluate the numerators.}\\\frac{2}{2},\frac{4}{2}\\\text{Simplify.}\\1,2\\\text{The midpoint of (-3, 5) and (5, -1) is (1, 2).}$

$\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\\\text{Substitute the values.}\\\frac{-4+6}{2},\frac{-3+3}{2}\\\text{Evaluate the numerators.}\\\frac{2}{2},\frac{0}{2}\\\text{Simplify.}\\1,0\\\text{The midpoint of (-4, -3) and (6, 3) is (1, 0).}$

3 0

3 years ago