Mathematics, 28.03.2020 21:36, school4life110

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

Polynomial Not a polynomial

Drag the expressions into the boxes to correctly complete the table. Polynomial Not a polynomial

Answers: 3

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answered: natalie2sheffield

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:

= non-negative integer

= 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

answered: Guest

Rounding decimals is very similar to rounding other numbers. if the hundredths and thousandths places of a decimal is forty-nine or less, they are dropped and the tenths place does not change. for example, rounding 0.843 to the nearest tenth would give 0.8. , your welcome !
What does x=? round to the nearest tenth

What does x=? round to the nearest tenth

answered: Guest

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.