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3 years ago

73210 the digit 3 stands

Mathematics

2 answers:

Nimfa-mama [501]3 years ago

5 0

The digit stands for 3,000.

Minchanka [31]3 years ago

3 0

To digit 3 stands for 3000

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This is the last part, you don't have to do number (6). I really appreciate the help so far!!!

marysya [2.9K]

Answer: just saying i realy need points or i cant ask questions anyways 4+x/-20=-40 is prob 45 or somthing

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2 years ago

Please answer 1-4 please

Nady [450]

Answer:

1. 1st Quadrant

2. 3rd Quadrant

3. 2nd Quadrant

4. 2nd Quadrant

5. 4rth Quadrant

6. 2nd Quadrant

7. No they do no correspond to the same point.

Step-by-step explanation:

1st Quadrant : ( positive , positive )

2nd Quadrant : ( negative , positive )

3rd Quadrant : ( negative , negative )

4th Quadrant : ( positive , negative )

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3 years ago

Read 2 more answers

Please help me with this!! I will mark brainlist!!

kondor19780726 [428]

Answer:

54°

Step-by-step explanation:

Your triangle is isosceles. You have been given the third angle, and you have been asked to find the measure of one of the two equal angles. We know that the internal angles of a triangle sum to 180, so $72+x+x=180 \rightarrow 2x=108 \rightarrow x = 54$

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3 years ago

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

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3 years ago

Solve log(2)(4x+8)=5

marysya [2.9K]

4x+8 = 32, 4x = 24, x = 6

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3 years ago