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asambeis [7]
3 years ago
5

Rationalize

tle="1 \div \sqrt{}3 - \sqrt{2} " alt="1 \div \sqrt{}3 - \sqrt{2} " align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
Marat540 [252]3 years ago
8 0

Answer:

1 \div \sqrt{}3 - \sqrt{2}

\frac{1 -  \sqrt{2} \sqrt{3}  }{ \sqrt{3} }

\frac{(1 -  \sqrt{6}) \sqrt{3}  }{ \sqrt{3} \sqrt{3}  }

\frac{ \sqrt{3}  -  \sqrt{6 \times 3} }{3}

\frac{ \sqrt{3} -  3\sqrt{2}  }{3}

is a required answer.

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Please help ASAP and explain the answer, thank you
jonny [76]

Answer:

Step-by-step explanation:

1) x² - 8x + 5 = 0

2) No subtract 5 from both sides

x² - 8x + 5 - 5 = 0 - 5

x² - 8x  = -5

3) Now find (8/2) = 4. Add 4² = 16 to both sides

x² - 8x + 16 = -5 + 16

x² - 8x + 16 = 11

4) (x - 4)²  = 11

5) Take square root both sides

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6 0
2 years ago
Write the equation of the line in fully simplified slope-intercept form.
Bad White [126]

Answer:

y = -\frac{2}{5} x-6

Step-by-step explanation:

1) First, find the slope of the line. Use the slope formula m =\frac{y_2-y_1}{x_2-x_1}. Pick two points on the line and substitute their x and y values into the formula, then solve. I used the points (-5,-4) and (0,-6):

m = \frac{(-6)-(-4)}{(0)-(-5)} \\m = \frac{-6+4}{0+5} \\m = \frac{-2}{5} \\  

So, the slope of the line is -\frac{2}{5}.

2) Next, use the point-slope formula y-y_1 = m (x-x_1) to write the equation of the line in point-slope form. (From there, we can convert it to slope-intercept form.) Substitute values for the m, x_1 and y_1 into the formula.

Since m represents the slope, substitute -\frac{2}{5}  in its place. Since x_1 and y_1 represent the x and y values of one point on the line, pick any point on the line (any one is fine, it will equal the same thing at the end) and substitute its x and y values in those places. (I chose (0,-6), as seen below.) Then, with the resulting equation, isolate y to put the equation in slope-intercept form:

y-(-6) = -\frac{2}{5} (x-(0))\\y + 6 = -\frac{2}{5} x\\y = -\frac{2}{5} x-6

6 0
3 years ago
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