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Ghella [55]
4 years ago
8

Pls help me wt this i need to do this by today

Mathematics
2 answers:
vovangra [49]4 years ago
7 0
59/6 is approx. 9.83 rounded to the hundredths.

the second one is 102.76.

the last one is 27.26.
Mashcka [7]4 years ago
6 0

Answer:

2.) 9.83

3.) 102.76

4.) 27.26

I hope this helps!!!

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<img src="https://tex.z-dn.net/?f=1%2F2x%5E3%20%2B%20x%20-7%20%3D%20-3%5Csqrt%7Bx%7D%20%20-1" id="TexFormula1" title="1/2x^3 + x
Margaret [11]

Answer:

  none of the above

Step-by-step explanation:

The problem as written cannot have any of the solutions offered.

For any of those choices, the right side expression will be irrational. The left side expression will be rational for any rational value of x, so cannot be equal to the right-side expression.

The solution is an irrational number near ...

  x ≈ 1.33682898582

5 0
4 years ago
Question:4x²-9 <br>answer:​
statuscvo [17]

Step-by-step explanation:

4x^2-9

=(2x)^2 -3^2

=(2x+3) (2x-3)

Hope this helps you.

3 0
3 years ago
P(x)= 3x^3-5x^2-14x-4
nexus9112 [7]
   
\displaystyle\\&#10;P(x)=3x^3-5x^2-14x-4\\\\&#10;D_{-4}=\{-4;~-2;~\underline{\bf -1};~1;~2;~4\}\\\\&#10;\text{We observe that } \frac{-1}{3} \text{ is a solution of the equation:}\\&#10;3x^3-5x^2-14x-4=0\\\\&#10;

\displaystyle\\&#10;\text{Verification}\\\\&#10;3x^3-5x^2-14x-4=\\\\&#10;=3\times\Big(-\frac{1}{3}\Big)^3-5\times\Big(-\frac{1}{3}\Big)^2-14\times\Big(-\frac{1}{3}\Big)-4=\\\\&#10;=-\frac{1}{9}-\frac{5}{9}+\frac{14}{3}-4=\\\\&#10;=-\frac{6}{9}+\frac{14}{3}-4=\\\\&#10;=-\frac{6}{9}+\frac{42}{9}- \frac{4\times 9}{9}=\\\\&#10; =-\frac{6}{9}+\frac{42}{9}- \frac{36}{9}= \frac{42-6-36}{9}=\frac{42-42}{9}=\frac{0}{9}=0\\\\&#10;\Longrightarrow~~~P(x)~\vdots~\Big(x+ \frac{1}{3}\Big)\\\\&#10;\Longrightarrow~~~P(x)~\vdots~(3x+1)


\displaystyle\\&#10;3x^3-5x^2-14x-4=0\\&#10;~~~~~-5x^2 = x^2 - 6x^2\\&#10;~~~~~-14x =-2x-12x \\&#10;3x^3+x^2 - 6x^2-2x-12x-4=0\\&#10;x^2(3x+1)-2x(3x+1) -4(3x+1)=0\\&#10;(3x+1)(x^2-2x -4)=0\\\\&#10;\text{Solve: } x^2-2x -4=0\\\\&#10;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}\\\\&#10;x_1 =1+\sqrt{5}\\&#10;x_2 =1-\sqrt{5}\\&#10;\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
If x is an even integer and y is an odd integer than x + y is a rational number. A) Always True B) Sometimes True C) Usually Tru
Katyanochek1 [597]
A) Always TRue
Both integers are rational, so the sum must be rational
7 0
4 years ago
Cos72cos130+sin72sin130
Setler [38]

Answer:

.5299192646

Step-by-step explanation:

Just plug it in a calculator

3 0
3 years ago
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