Ask question

Ask question

All categories

4 years ago

12

Which algebraic expression is a polynomial? 4x2 – 3x + StartFraction 2 Over x EndFraction –6x3 + x2 – StartRoot 5 EndRoot 8x2 +

StartRoot x EndRoot –2x4 + StartFraction 3 Over 2 x EndFraction

2 answers:

rusak2 [61]4 years ago

7 0

Answer:

The Answer is B

Step-by-step explanation:

Veseljchak [2.6K]4 years ago

3 0

Answer:

B. $-6x^3+x^2-\sqrt5$

Step-by-step explanation:

Polynomial of one variable:It is that expression in which power of each variable in the given should be a whole number.The highest power of variable is called degree of polynomial.

Power of variable in given terms cannot be negative or fraction.

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

It can be write as

$4x^2-3x+2x^{-1}$

By definition of polynomial of one variable , it is not a polynomial.

B. $-6x^3+x^2-\sqrt5$

By definition of polynomial of one variable , it is a polynomial.

C. $8x^2+\sqrt{x}$

By definition of polynomial of one variable, it is not a polynomial.

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

By definition of polynomial of one variable,it is not a polynomial.

Hence, option B is true.

You might be interested in

Pleaseeeee ı dont find the answer

mote1985 [20]

Answer: $y(x) = \sqrt{\frac{7x^{14}}{-2x^7+9}}\\\\$

==========================================================

Explanation:

The given differential equation (DE) is

$y'-\frac{7}{x}y = \frac{y^3}{x^8}\\\\$

Which is the same as

$y'-\frac{7}{x}y = \frac{1}{x^8}y^3\\\\$

This 2nd DE is in the form $y' + P(x)y = Q(x)y^n$

where

$P(x) = -\frac{7}{x}\\\\Q(x) = \frac{1}{x^8}\\\\n = 3$

As the instructions state, we'll use the substitution $u = y^{1-n}$

We specifically use $u = y^{1-n} = y^{1-3} = y^{-2}$

-----------------

After making the substitution, we'll end up with this form

$\frac{du}{dx} + (1-n)P(x)u = (1-n)Q(x)\\\\$

Plugging in the items mentioned, we get:

$\frac{du}{dx} + (1-n)P(x)u = (1-n)Q(x)\\\\\frac{du}{dx} + (1-3)*\frac{-7}{x}u = (1-3)\frac{1}{x^8}\\\\\frac{du}{dx} + \frac{14}{x}u = -\frac{2}{x^8}\\\\$

We can see that we have a new P(x) and Q(x)

$P(x) = \frac{14}{x}\\\\Q(x) = -\frac{2}{x^8}$

-------------------

To solve the linear DE $\frac{du}{dx} + \frac{14}{x}u = -\frac{2}{x^8}\\\\$ , we'll need the integrating factor which I'll call m

$m(x) = e^{\int P(x) dx} = e^{\int \frac{14}{x}dx} = e^{14\ln(x)}$

$m(x) = e^{\ln(x^{14})} = x^{14}$

We will multiply both sides of the linear DE by this m(x) integrating factor to help with further integration down the road.

$\frac{du}{dx} + \frac{14}{x}u = -\frac{2}{x^8}\\\\m(x)*\left(\frac{du}{dx} + \frac{14}{x}u\right) = m(x)*\left(-\frac{2}{x^8}\right)\\\\x^{14}*\frac{du}{dx} + x^{14}*\frac{14}{x}u = x^{14}*\left(-\frac{2}{x^8}\right)\\\\x^{14}*\frac{du}{dx} + 14x^{13}*u = -2x^6\\\\\left(x^{14}*u\right)' = -2x^6\\\\$

It might help to think of the product rule being done in reverse.

Now we can integrate both sides to solve for u

$\left(x^{14}*u\right)' = -2x^6\\\\\displaystyle \int\left(x^{14}*u\right)'dx = \int -2x^6 dx\\\\\displaystyle x^{14}*u = \frac{-2x^7}{7}+C\\\\\displaystyle u = x^{-14}*\left(\frac{-2x^7}{7}+C\right)\\\\\displaystyle u = x^{-14}*\frac{-2x^7}{7}+Cx^{-14}\\\\\displaystyle u = \frac{-2x^{-7}}{7}+Cx^{-14}\\\\$

$u = \frac{-2}{7x^7} + \frac{C}{x^{14}}\\\\u = \frac{-2}{7x^7}*\frac{x^7}{x^7} + \frac{C}{x^{14}}*\frac{7}{7}\\\\u = \frac{-2x^7}{7x^{14}} + \frac{7C}{7x^{14}}\\\\u = \frac{-2x^7+7C}{7x^{14}}\\\\$

Unfortunately, this isn't the last step. We still need to find y.

Recall that we found $u = y^{-2}\\\\$

So,

$u = \frac{-2x^7+7C}{7x^{14}}\\\\y^{-2} = \frac{-2x^7+7C}{7x^{14}}\\\\y^{2} = \frac{7x^{14}}{-2x^7+7C}$

We're told that y(1) = 1. This means plugging x = 1 leads to the output y = 1. So the RHS of the last equation should lead to 1. We'll plug x = 1 into that RHS, set the result equal to 1 and solve for C

$\frac{7*1^{14}}{-2*1^7+7C} = 1\\\\\frac{7}{-2+7C} = 1\\\\7 = -2+7C\\\\7+2 = 7C\\\\7C = 9\\\\C = \frac{9}{7}$

So,

$y^{2} = \frac{7x^{14}}{-2x^7+7C}\\\\y^{2} = \frac{7x^{14}}{-2x^7+7*\frac{9}{7}}\\\\y^{2} = \frac{7x^{14}}{-2x^7+9}\\\\y = \sqrt{\frac{7x^{14}}{-2x^7+9}}\\\\$

We go with the positive version of the root because y(1) is positive, which must mean y(x) is positive for all x in the domain.

3 0

3 years ago

Participants enter a research study with unique characteristics that produce different scores from one person to another. For an

agasfer [191]

Answer:

check online for more information

5 0

3 years ago

Bob was given an allowance of 115 dollars. He spent 15% of it on food, and 60% of the remainder on games. How much money did Bo

WINSTONCH [101]

Hello I hope this makes it easier for you to understand.
All you have to do is find ”What is 60% of 115?”

STEP 1: Multiply 115 by 0.6 (60%)

STEP 2: = $69

Bob spent $69 on games.

Hope this helps.

4 0

3 years ago

P-2 times what = 4-p^2

tino4ka555 [31]

Answer:

-2-p

Step-by-step explanation:

(p-2)x=4-p^2

(4-p^2)/(p-2)

-2-p

4 0

3 years ago

Read 2 more answers

Write an equation of that passes through the points (-2,-3,) and (2,5)

MAVERICK [17]

Answer:

y = 2x + 1

Step-by-step explanation:

Find the slope

( -2 , -3) ( 2 , 5)

m = (y2 - y1 )/(x2 - x1)

x1= -2

y1 = -3

x2 = 2

y2 = 5

m = 5 - (-3)/ ( 2 - (-2)

m = (5 + 3) / ( 2 + 2)

= 8/4

= 2

m = 2

Using the equation of a line

y = mx + c

y - intercept point y

m - slope of the line

x - intercept point x

c - intercept

Substitute m into the equation

y = 2x + c

Substitute any of the two points given into the equation

( 2 ,5 )

x - 2

y - 5

y = 2x + c

5 = 2(2) + c.

5 = 4 + c

c = 5 - 4

c = 1

The equation of the line

y = 2x + 1

5 0

3 years ago