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3 years ago

How can you quickly determine the number of roots a polynomial will have by looking at the equation?

2 answers:

8 0

Suppose we have a generic polynomial of the form:
$ax ^ 2 + bx + c
$
To know how many roots the polynomial can have, the first thing you should do is observe the term of greatest exponent.
For this case, the term of greatest exponent is 2.
Therefore, the polynomial has 2 roots.
Answer:
You must observe the term of the polynomial with greater exponent.

34kurt3 years ago

5 0

The degree of polynomial is equivalent to the number of roots a polynomial has.

The Linear Factorization Theorem states that a polynomial of degree n has exactly n linear factors, and each can be written in the form (x - c) , where c is a root.

There are two types of roots. The real root and the complex root.

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How much should you invest at 4.8% compounded continuously to have $5000 in 2 years?

mezya [45]

Answer:

The total compound interest is $502.74 i think

Step-by-step explanation:

7 0

3 years ago

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In ΔGHI, the measure of ∠I=90°, the measure of ∠G=82°, and GH = 3.4 feet. Find the length of HI to the nearest tenth of a foot.

taurus [48]

Answer:

3.4

Step-by-step explanation:

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3 years ago

Read 2 more answers

Drag the expressions into the boxes to correctly complete the table.

lora16 [44]

Answer:

SUMMARY:

$x^4+\frac{5}{x^3}-\sqrt{x}+8$ → Not a Polynomial

$-x^5+7x-\frac{1}{2}x^2+9$ → A Polynomial

$x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi$ → A Polynomial

$\left|x\right|^2+4\sqrt{x}-2$ → Not a Polynomial

$x^3-4x-3$ → A Polynomial

$\frac{4}{x^2-4x+3}$ → Not a Polynomial

Step-by-step explanation:

The algebraic expressions are said to be the polynomials in one variable which consist of terms in the form $ax^n$ .

Here:

$n$ = non-negative integer

$a$ = is a real number (also the the coefficient of the term).

Lets check whether the Algebraic Expression are polynomials or not.

Given the expression

$x^4+\frac{5}{x^3}-\sqrt{x}+8$

If an algebraic expression contains a radical in it then it isn’t a polynomial. In the given algebraic expression contains $\sqrt{x}$ , so it is not a polynomial.

Also it contains the term $\frac{5}{x^3}$ which can be written as $5x^{-3}$ , meaning this algebraic expression really has a negative exponent in it which is not allowed. Therefore, the expression $x^4+\frac{5}{x^3}-\sqrt{x}+8$ is not a polynomial.

Given the expression

$-x^5+7x-\frac{1}{2}x^2+9$

This algebraic expression is a polynomial. The degree of a polynomial in one variable is considered to be the largest power in the polynomial. Therefore, the algebraic expression is a polynomial is a polynomial with degree 5.

Given the expression

$x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi$

in a polynomial with a degree 4. Notice, the coefficient of the term can be in radical. No issue!

Given the expression

$\left|x\right|^2+4\sqrt{x}-2$

is not a polynomial because algebraic expression contains a radical in it.

Given the expression

$x^3-4x-3$

a polynomial with a degree 3. As it does not violate any condition as mentioned above.

Given the expression

$\frac{4}{x^2-4x+3}$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}$

Therefore, is not a polynomial because algebraic expression really has a negative exponent in it which is not allowed.

SUMMARY:

$x^4+\frac{5}{x^3}-\sqrt{x}+8$ → Not a Polynomial

$-x^5+7x-\frac{1}{2}x^2+9$ → A Polynomial

$x^4+x^3\sqrt{7}+2x^2-\frac{\sqrt{3}}{2}x+\pi$ → A Polynomial

$\left|x\right|^2+4\sqrt{x}-2$ → Not a Polynomial

$x^3-4x-3$ → A Polynomial

$\frac{4}{x^2-4x+3}$ → Not a Polynomial

3 0

3 years ago

What is the slope of the line that passes through points (1/4,3) and (5,8)

Alex73 [517]

Answer:

20/19

Step-by-step explanation:

y2-y1/x2-x1

= 8-3/5-1/4

=20/19

8 0

3 years ago

Can some one help me on this pls :(

Julli [10]

Answer:

Step-by-step explanation:

1= 2/3

2=5/6

3=21/40

4= 2 11/20

5= 1 1/3

6=5/6

7=1 1/2

8=9/10

9= 5 1/6

10= 2 1/2

6 0

2 years ago