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3 years ago

Write the equation of the perpendicular bisector of the line segment connecting (10, 3) to (6, -13)

Mathematics

1 answer:

Feliz [49]3 years ago

3 0

Answer:

y = -1/4x - 3

Step-by-step explanation:

Halfway through would be:

(10 + 6 / 2, -13 + 3 / 2) --> (8, -5)

The slope of the line that goes through (10, 3) and (6, -13) is -13 - 3 / 6 - 10 = -16 / -4 = 4.

That means the slope of the line that is perpendicular would be the negative reciprocal of 4, which is -1/4.

Put this in point slope form with slope = -1/4 and point = (8, -5):

y + 5 = -1/4(x - 8)

y + 5 = -1/4x + 2

y = -1/4x - 3

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- Polynomial Functions -For each function, state the vertex; whether the vertex is a maximum or minimum point; the equation of t

vladimir2022 [97]

EXPLANATION

Given the function f(x) = (x-6)^2 + 1

$\mathrm{The\: vertex\: of\: an\: up-down\: facing\: parabola\: of\: the\: form}\: y=ax^2+bx+c\: \mathrm{is}\: x_v=-\frac{b}{2a}$

Expanding (x-6)^2 + 1 by applying the Perfect Square Formula:

$=x^2-12x+37$ $\mathrm{The\: parabola\: params\: are\colon}$ $a=1,\: b=-12,\: c=37$ $x_v=-\frac{b}{2a}$ $x_v=-\frac{\left(-12\right)}{2\cdot\:1}$ $\mathrm{Simplify}$ $x_v=6$ $y_v=6^2-12\cdot\: 6+37$

Simplify:

$y_v=1$

$\mathrm{Therefore\: the\: parabola\: vertex\: is}$ $\mleft(6,\: 1\mright)$ $\mathrm{If}\: a$ $\mathrm{If}\: a>0,\: \mathrm{then\: the\: vertex\: is\: a\: minimum\: value}$ $a=1$ $\mathrm{Minimum}\mleft(6,\: 1\mright)$ $\mathrm{For\: a\: parabola\: in\: standard\: form}\: y=ax^2+bx+c\: \mathrm{the\: axis\: of\: symmetry\: is\: the\: vertical\: line\: that\: goes\: through\: the\: vertex}\: x=\frac{-b}{2a}$

Expanding (x-6)^2 + 1 by applying the Perfect Square Formula:

$y=x^2-12x+37$ $\mathrm{Axis\: of\: Symmetry\: for}\: y=ax^2+bx+c\: \mathrm{is}\: x=\frac{-b}{2a}$ $a=1,\: b=-12$ $x=\frac{-\left(-12\right)}{2\cdot\:1}$ $\mathrm{Refine}$

Axis of simmetry : x=6

The quadratic function has the same shape than the parent function y=x^2 because there is NOT a coefficient within x.

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1 year ago

Please help as soon as possible

Kobotan [32]

We have an triangle:
base=4 in
height=3 in,
This triangle can be dividided into two equal triangles, we need calculate the hypotenuse.
leg₁=4 in/2=2 in
leg₂=3 in

Pythagoras law:
hypotenuse²=leg₁²+leg₂²
hypotenuse²=(2 in)²+(3 in)²
hypotenuse²=4 in²+9 in²
hypotenuse²=13 in²
hypotenuse=√13 in.

Now, we can find the surface area.
Surface area=2 *(rectangle area)+base area + 2(triangle area)
rectangle area=10 in x √13 in=10√13 in²
base area=10 in x 4 in=40 in²
Triangle area=(4 in x 3 in)/2=6 in²

Surface area=2(10√13 in²)+40 in²+2(6 in²)=(20√13+52) in²≈124.11 in²

Answer: 124.11 in²

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