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Ivenika [448]
3 years ago
14

Is the relation a function? Explain Set-builder notation? Domain/range?

Mathematics
2 answers:
saveliy_v [14]3 years ago
4 0

Answer:

doamin = set builder nation

bija089 [108]3 years ago
3 0

Answer:

domain= set builder notation

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What's an equation for "Six more then twice a number is 72?"
Tema [17]

the equation would be:

2x+6 = 72

6 0
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In need help to answer this question ​
andrezito [222]

Answer:

1 6/7

Step-by-step explanation:

1 & 6/7

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P(x)= 3x^3-5x^2-14x-4
nexus9112 [7]
   
\displaystyle\\
P(x)=3x^3-5x^2-14x-4\\\\
D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\
\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\
3x^3-5x^2-14x-4=0\\\\


\displaystyle\\
\text{Verification}\\\\
3x^3-5x^2-14x-4=\\\\
=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\
=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{14}{3}-4=\\\\
=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\
 =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\
\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\
\Longrightarrow~~~P(x)~\vdots~(3x+1)


\displaystyle\\
3x^3-5x^2-14x-4=0\\
~~~~~-5x^2 = x^2 - 6x^2\\
~~~~~-14x =-2x-12x \\
3x^3+x^2 - 6x^2-2x-12x-4=0\\
x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\
(3x+1)(x^2-2x -4)=0\\\\
\text{Solve: } x^2-2x -4=0\\\\
x_{12}= \frac{-b\pm  \sqrt{b^2-4ac}}{2a}=\\\\=\frac{2\pm  \sqrt{4+16}}{2}=\frac{2\pm  \sqrt{20}}{2}=\frac{2\pm  2\sqrt{5}}{2}=1\pm\sqrt{5}\\\\
x_1 =1+\sqrt{5}\\
x_2 =1-\sqrt{5}\\
\Longrightarrow P(x)= 3x^3-5x^2-14x-4 =\boxed{(3x+1)(x-1-\sqrt{5})(x-1+\sqrt{5})}



7 0
3 years ago
Which number has the same value
photoshop1234 [79]
Answer:
D 0.08

Explanation:
dunno how to
7 0
3 years ago
Solve the quadratic.<br><br><br> 4x^2+4-20=0<br><br><br><br><br> brainlist if you explain
frez [133]

Answer:

  • x = -1/2(1 +√21) ≈ -2.79129
  • x = -1/2(1 -√21) ≈ 1.79129

Step-by-step explanation:

We assume the middle term is supposed to be 4x.

We can remove a common factor of 4 to simplify this a bit.

  x^2 +x -5 = 0

This is of the form

  ax^2 +bx +c = 0

where a=1, b=1, c=-5.

__

The <em>quadratic formula</em> gives the solutions as ...

  x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}

Filling in the given coefficients, we have ...

  x = (-1 ±√(1^2 -4·1·(-5)))/(2·1)

  x = (-1±√21)/2

The solutions are x = -1/2(1 +√21) and -1/2(1 -√21).

_____

<em>If what you wrote is what you intend</em>, then the equation simplifies to 4x^2 -16 = 0.

Dividing by 4 and factoring the difference of squares gives ...

  x^2 -4 = 0

  (x -2)(x +2) = 0

These factors are zero (hence their product is 0) for the values x = 2 and x = -2.

The solutions are x=2 and x=-2.

5 0
3 years ago
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