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2 years ago

40 QUESTIONS FOR JUST ONE QUESTION :) QUESTION 6

Mathematics

1 answer:

7nadin3 [17]2 years ago

3 0

Answer: <em>Credit to work in process invetory-assembly</em>

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Big freh points to da butter dog da dog wit da butter

Doss [256]

Answer:

Budda Dawgggg

Step-by-step explanation:

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2 years ago

What is the relationship between 5.34×1055.34×105 and 5.34×10−25.34×10−2 ? Select from the drop-down menus to correctly complete

Anton [14]

The relationship between 5.34×105 and 5.34×10−2 would be that the former number is 10000000 times larger than the latter value. We can obtain this by dividing 5.34×105 and 5.34×10−2 where the quotient would be 10000000. Hope this answers the question. Have a nice day.

5 0

3 years ago

Read 2 more answers

Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

3 0

2 years ago

Please need help on this question.

Stella [2.4K]

Answer:

(4,5)

Step-by-step explanation:

8 0

3 years ago

1/x-2 graphed<br><br><img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx-2%7D" id="TexFormula1" title="\frac{1}{x-2}" alt="\fra

bearhunter [10]

Answer:

Step-by-step explanation:

The given equation is expressed as

y = 1/(x - 2)

We will assume values for x and determine the corresponding values for y. Therefore,

When x = 0, y = 1/(0 - 2) = - 1/2

When x = 1, y = 1/(1 - 2) = - 1

When x = 2, y = 1/(2 - 2) = 0

When x = 3, y = 1/(3 - 2) = 1

When x = 4, y = 1/(4 - 2) = 1/2

To plot the graph, we would choose a suitable scale for the x and y axis of the graph.

Let 2 cm represent 1 unit on the vertical axis.

Let 2 cm represent 0.5 unit on the x axis.

The graph is shown in the attached photo

4 0

3 years ago