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stellarik [79]
3 years ago
11

If several points are graphed on a number line, is the point that is the farthest from 0 always the greatest?

Mathematics
2 answers:
Kryger [21]3 years ago
8 0

Answer:

If several points are graphed on a number line, the point that is farthest from 0 need NOT be always greater. Consider x axis, our Real number line. We can traverse in positive number direction and negative number direction.

mark me at brainlist

KengaRu [80]3 years ago
8 0

Answer:

No

Step-by-step explanation:

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Answer:

(-4,4)

Step-by-step explanation:

-5x-10y=-20

10x+10y=0

divide both sides by 2

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5 0
3 years ago
Find all solutions to
BARSIC [14]

Answer:

x= 0 , \frac{1}{14} , \frac{-1}{12}

Step-by-step explanation:

Given, equation is \sqrt[3]{15x-1} + \sqrt[3]{13x+1} = 4\sqrt[3]{x}. →→→ (1)

Now, by cubing the equation on both sides, we get

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(since from (1),  \sqrt[3]{15x-1} + \sqrt[3]{13x+1} = 4\sqrt[3]{x})

⇒ 12× \sqrt[3]{15x-1}×\sqrt[3]{13x+1} (\sqrt[3]{x})= 36x.

⇒ 3x = \sqrt[3]{(15x-1)(13+1)(x)}.

Now, once again cubing on both sides, we get

(3x)³ = (\sqrt[3]{(15x-1)(13+1)(x)})³.

⇒ 27x³ = (15x-1)(13x+1)(x).

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⇒ by, solving the equation we get ,

x = 0 ; x = \frac{1}{14} ; x = \frac{-1}{12}

therefore, solution is x= 0 , \frac{1}{14} , \frac{-1}{12}

7 0
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$5500×1.12= $6160. same method as the previous answer I gave you:)
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