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Temka [501]
2 years ago
8

Point Z is the circumcenter of ΔLMN.

Mathematics
1 answer:
vovikov84 [41]2 years ago
5 0

Answer: Option D.

Step-by-step explanation: I did this.

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Whats 4/1-4 - 7/3 to its reduced form
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54

Step-by-step explanation:

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How many points will I get if I get 60 points every 5 mins and it goes on for 4 hours?
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Someone PLEASE HELP ME<br><br>Find the slope of the line <br><br>Algebra 1
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Step-by-step explanation:

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3 years ago
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1101101+11011 in binary numbers please help
wel

Answer:

The answer is 1000 1000

Step-by-step explanation:

In HEX, it would be 88.

In DEC, it would be 136.

In Oct, it would be 210.

Binary, how to add numbers together.

     1101101

+       11011

10001000

This is just like adding regular numbers together.

For example, when you add 88 to 55, this is what will happen

   1(r)

  88

+ 55

13(+1)  3

143

With binary, when 1 and 0 are added together, the result is 1. When 1 and 1 are added together, the result is 1 and 0 (1 being the "remainder", or the number that is carried over. This number is then added to the next part of the number when adding, just like in normal addition.

7 0
2 years ago
Solve irrational equation pls
rusak2 [61]
\hbox{Domain:}\\&#10;x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\&#10;x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\&#10;x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\&#10;(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\&#10;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)\\&#10;x\in(-\infty,-2\rangle\cup\langle3,\infty)


&#10;\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\&#10;x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\&#10;2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\&#10;\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\&#10;(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\&#10;(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\&#10;(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\&#10;(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\&#10;(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\&#10;4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\
&#10;4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\&#10;(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\&#10;(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\&#10;(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\&#10;(x-1)^2(3x^2-28)=0\\&#10;x-1=0 \vee 3x^2-28=0\\&#10;x=1 \vee 3x^2=28\\&#10;x=1 \vee x^2=\dfrac{28}{3}\\&#10;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\\&#10;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}}\}\\&#10;\boxed{\boxed{x=1}}
3 0
2 years ago
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