Point T(-9,5) lies on the perpendicular bisector of UV. If the

Mathematics

1 answer:

Flura [38]3 years ago

5 0

The coordinate of point V is at (-16, 9)

If Point T(-9,5) lies on the perpendicular bisector of UV, this means that the point divides the line UV into two equal parts

Given the following coordinates

Midpoint T = (-9, 5)

U = (-2, 1)

Required

coordinate of point V

Using the midpoint formulas;

$T(x, y) = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) \\T(-9, 5) = (\frac{-2+x_2}{2}, \frac{1+y_2}{2})$

Get the value if x₂ and y₂

$-9 = \frac{-2+x_2}{2}\\-18 = -2+x_2\\x_2 = -18+2\\x_2 = -16\\$

Similarly;

$5 = \frac{1+y_2}{2}\\10 = 1+y_2\\y_2 = 10-1\\y_2 = 9\\$

Hence the coordinate of point V is at (-16, 9)

Learn more here: brainly.com/question/18049211

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Mathematics 13 don’t understand

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Answer:

1/6^3

Step-by-step explanation:

The applicable rules of exponents are ...

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Your expression can be simplified as follows:

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<em>Additional comment</em>

If you think of an exponent as signifying repeated multiplication, the rules of exponents may be easier to remember. The exponent tells you how many times the base is a factor in the product.

Consider multiplication:

$(x\cdot x\cdot x)\cdot(x\cdot x)=x^3\cdot x^2=x^{3+2}=x^5\\\\(x\cdot x\cdot x)\cdot(x\cdot x\cdot x)=(x^3)^2=x^{3\cdot2}=x^6$

Consider division:

$\dfrac{x\cdot x\cdot x}{x\cdot x}=x\quad\Longleftrightarrow\quad\dfrac{x^3}{x^2}=x^{3-2}=x^1\\\\\dfrac{x\cdot x}{x\cdot x\cdot x}=\dfrac{1}{x}\quad\Longleftrightarrow\quad\dfrac{x^2}{x^3}=x^{2-3}=x^{-1}$