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Kobotan [32]
3 years ago
15

Pls HELPPP, BRAINLIEST for best and right answer

Mathematics
2 answers:
DerKrebs [107]3 years ago
8 0
The numbers that make it true is 4 6 and 7. just put the numbers in for r and if you put in 4 it would just be 10 which isn’t greater OR equal to 11
xxMikexx [17]3 years ago
5 0

Answer:

i think the answer is 5

Step-by-step explanation:

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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
Please help u don't get it #begging
Jobisdone [24]
33.C
34.C
Those are the answers
4 0
3 years ago
Find the inverse: f(x)=x+5/3x-1<br><br> f^-1x=
babunello [35]

We have f(x)=\dfrac{x+5}{3x-1}.

To find inverse function f^{-1}(x) we substitute x with f^{-1}(x) and vice-versa to get

x=\dfrac{f^{-1}(x)+5}{3f^{-1}(x)-1}

Now solve for f^{-1}(x). Note that I will use j instead.

x=\dfrac{j+5}{3j-1} \\x(3j-1)=j+5 \\3jx-x=j+5 \\3jx-x-j-5=0 \\3jx-j=x+5 \\j(3x-1)=x+5 \\j=\dfrac{x+5}{3x-1}

So we find that f(x)=f^{-1}.

Hope this helps.

7 0
3 years ago
Read 2 more answers
Identify the fraction that is equivalent to 4/7
damaskus [11]
4/7,8/14,12/21,16/28,20/35,24/42,28/49,32/56,36/63,40/70 and so on are equivalent fractions to 4/7
4 0
3 years ago
Read 2 more answers
Evaluate 2r (t – 1) ifr =<br> 4 and t<br> = 6
garri49 [273]

Answer:

40

Step-by-step explanation:

Step 1: Define

2r(t - 1)

r = 4

t = 6

Step 2: Substitute and Evaluate

2(4) · (6 - 1)

8 · 5

40

4 0
3 years ago
Read 2 more answers
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