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svet-max [94.6K]
2 years ago
13

A(3,4) and B(-3,2) are pointd on a coordinate plane. find the coordinate of a points C on the x axis such that AC=BC​

Mathematics
1 answer:
mr Goodwill [35]2 years ago
7 0

Answer:

Step-by-step explanation:

Here's the game plan. In order to find a point on the x-axis that makes AC = BC, we need to find the midpoint of AB and the slope of AB. From there, we can find the equation of the line that is perpendicular to AB so we can then fit a 0 in for y and solve for x. This final coordinate will be the answer you're looking for. First and foremost, the midpoint of AB:

and

Now for the slope of AB:

and

 So if the slope of AB is 1/3, then the slope of a line perpendicular to that line is -3. What we are finding now is the equation of the line perpendicular to AB and going through (0, 3):

and filling in:

y - 3 = -3(x - 0) and

y - 3 = -3x + 0 and

y - 3 = -3x so

y = -3x + 3. Filling in a 0 for y will give us the coordinate we want for the x-intercept (the point where this line goes through the x-axis):

0 = -3x + 3 and

-3 = -3x so

x = 1

The coordinate on the x-axis such that AC = BC is (1, 0)

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7nadin3 [17]

Answer:

g(x) = x² - 4 is already in form of a variable, I.e., x

g(4x) takes another variable, I.e., 4x

Same as before, 4x takes over x:

=> g(4x) = (4x)² - 4

  • <em>(</em><em>ax</em><em>)</em><em>²</em><em> </em><em>=</em><em> </em><em>a</em><em>²</em><em>x</em><em>²</em><em>,</em><em> </em><em>where</em><em> </em><em>a</em><em> </em><em>is</em><em> </em><em>some</em><em> </em><em>arbitrary</em><em> </em><em>constant</em><em>.</em><em> </em>

<h3><u>Answer</u><u>:</u> </h3>

=> g(4x) = 16x² - 4

OR

=> g(4x) = 4{4x² - 1}

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2 years ago
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How to factor 9x^2+5x-12
bekas [8.4K]

Hi Stacey


9x^2+5x-12

Answer : Doesn't factor

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Determine the correct value of 3x + 2 when x = 5. Show your work
Alexxandr [17]

Answer:

17

Step-by-step explanation:

3 multiply by 5 (which is x) = 15+2

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Prove that the segments joining the midpoint of consecutive sides of an isosceles trapezoid form a rhombus.
sergiy2304 [10]

Answer:

See explanation

Step-by-step explanation:

a) To prove that DEFG is a rhombus, it is sufficient to prove that:

  1. All the sides of the rhombus are congruent:  |DG|\cong |GF| \cong |EF| \cong |DE|
  2. The diagonals are perpendicular

Using the distance formula; d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

|DG|=\sqrt{(0-(-a-b))^2+(0-c)^2}

\implies |DG|=\sqrt{a^2+b^2+c^2+2ab}

|GF|=\sqrt{((a+b)-0)^2+(c-0)^2}

\implies |GF|=\sqrt{a^2+b^2+c^2+2ab}

|EF|=\sqrt{((a+b)-0)^2+(c-2c)^2}

\implies |EF|=\sqrt{a^2+b^2+c^2+2ab}

|DE|=\sqrt{(0-(-a-b))^2+(2c-c)^2}

\implies |DE|=\sqrt{a^2+b^2+c^2+2ab}

Using the slope formula; m=\frac{y_2-y_1}{x_2-x_1}

The slope of EG is m_{EG}=\frac{2c-0}{0-0}

\implies m_{EG}=\frac{2c}{0}

The slope of EG is undefined hence it is a vertical line.

The slope of  DF is m_{DF}=\frac{c-c}{a+b-(-a-b)}

\implies m_{DF}=\frac{0}{2a+2b)}=0

The slope of DF is zero, hence it is a horizontal line.

A horizontal line meets a vertical line at 90 degrees.

Conclusion:

Since |DG|\cong |GF| \cong |EF| \cong |DE| and DF \perp FG , DEFG is a rhombus

b) Using the slope formula:

The slope of DE is m_{DE}=\frac{2c-c}{0-(-a-b)}

m_{DE}=\frac{c}{a+b)}

The slope of FG is m_{FG}=\frac{c-0}{a+b-0}

\implies m_{FG}=\frac{c}{a+b}

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iVinArrow [24]

Answe3.141592654

Step-by-step explanation:

By mupltjirini

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