E

∆ABC has vertices A(–2, 0), B(0, 8), and C(4, 2)

Natali [406]

Answer:

Part 1) The equation of the perpendicular bisector side AB is $y=-\frac{1}{4}x+\frac{15}{4}$

Part 2) The equation of the perpendicular bisector side BC is $y=\frac{2}{3}x+\frac{11}{3}$

Part 3) The equation of the perpendicular bisector side AC is $y=-3x+4$

Part 4) The coordinates of the point P(0.091,3.727)

Step-by-step explanation:

Part 1) Find the equation of the perpendicular bisector side AB

we have

A(–2, 0), B(0, 8)

step 1

Find the slope AB

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{8-0}{0+2}$

$m=4$

step 2

Find the slope of the perpendicular line to side AB

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

therefore

The slope is equal to

$m=-\frac{1}{4}$

step 3

Find the midpoint AB

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M(\frac{-2+0}{2},\frac{0+8}{2})$

$M(-1,4)$

step 4

Find the equation of the perpendicular bisectors of AB

the slope is $m=-\frac{1}{4}$

passes through the point $(-1,4)$

The equation in slope intercept form is equal to

$y=mx+b$

substitute

$4=(-\frac{1}{4})(-1)+b$

solve for b

$b=4-\frac{1}{4}$

$b=\frac{15}{4}$

so

$y=-\frac{1}{4}x+\frac{15}{4}$

Part 2) Find the equation of the perpendicular bisector side BC

we have

B(0, 8) and C(4, 2)

step 1

Find the slope BC

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{2-8}{4-0}$

$m=-\frac{3}{2}$

step 2

Find the slope of the perpendicular line to side BC

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

therefore

The slope is equal to

$m=\frac{2}{3}$

step 3

Find the midpoint BC

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M(\frac{0+4}{2},\frac{8+2}{2})$

$M(2,5)$

step 4

Find the equation of the perpendicular bisectors of BC

the slope is $m=\frac{2}{3}$

passes through the point $(2,5)$

The equation in slope intercept form is equal to

$y=mx+b$

substitute

$5=(\frac{2}{3})(2)+b$

solve for b

$b=5-\frac{4}{3}$

$b=\frac{11}{3}$

so

$y=\frac{2}{3}x+\frac{11}{3}$

Part 3) Find the equation of the perpendicular bisector side AC

we have

A(–2, 0) and C(4, 2)

step 1

Find the slope AC

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

substitute the values

$m=\frac{2-0}{4+2}$

$m=\frac{1}{3}$

step 2

Find the slope of the perpendicular line to side AC

Remember that

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

therefore

The slope is equal to

$m=-3$

step 3

Find the midpoint AC

The formula to calculate the midpoint between two points is equal to

$M(\frac{x1+x2}{2},\frac{y1+y2}{2})$

substitute the values

$M(\frac{-2+4}{2},\frac{0+2}{2})$

$M(1,1)$

step 4

Find the equation of the perpendicular bisectors of AC

the slope is $m=-3$

passes through the point $(1,1)$

The equation in slope intercept form is equal to

$y=mx+b$

substitute

$1=(-3)(1)+b$

solve for b

$b=1+3$

$b=4$

so

$y=-3x+4$

Part 4) Find the coordinates of the point of concurrency of the perpendicular bisectors (P)

we know that

The point of concurrency of the perpendicular bisectors is called the circumcenter.

Solve by graphing

using a graphing tool

the point of concurrency of the perpendicular bisectors is P(0.091,3.727)

see the attached figure

5 0

3 years ago

Elena and Jada are 12 miles apart on a path when they start moving towards each other. Elena answered a constant speed of 5 mile

djverab [1.8K]

Answer:

i think it is 234 i am not sure

Step-by-step explanation:

3 0

2 years ago

Totsakan school is selling tickets to a choral performance on the first day of ticket sales the school sold 10 senior tickets an

ohaa [14]

Given :

On the first day of ticket sales the school sold 10 senior tickets and 1 child ticket for a total of $85 .

The school took in $75 on the second day by selling 5 senior citizens tickets and 7 child tickets.

To Find :

The price of a senior ticket and the price of a child ticket.

Solution :

Let, price of senior ticket and child ticket is x and y respectively.

Mathematical equation of condition 1 :

10x + y = 85 ...1)

Mathematical equation of condition 2 :

5x + 7y = 75 ...2)

Solving equation 1 and 2, we get :

2(2) - (1) :

2( 5x + 7y - 75 ) - ( 10x +y - 85 ) = 0

10x + 14y - 150 - 10x - y + 85 = 0

13y = 65

y = 5

10x - 5 = 85

x = 8

Therefore, price of a senior ticket and the price of a child ticket $8 and $5.

Hence, this is the required solution.

5 0

3 years ago

Let w = x2 + y2 + z2, x = uv, y = u cos(v), z = u sin(v). use the chain rule to find ∂w ∂u when (u, v) = (9, 0).

jasenka [17]

By the chain rule,

$\dfrac{\partial w}{\partial u}=\dfrac{\partial w}{\partial x}\cdot\dfrac{\partial x}{\partial u}+\dfrac{\partial w}{\partial y}\cdot\dfrac{\partial y}{\partial u}+\dfrac{\partial w}{\partial z}\cdot\dfrac{\partial z}{\partial u}$
$\dfrac{\partial w}{\partial u}=2xv+2y\cos v+2z\sin v$

We also have $x(9,0)=0$ , $y(9,0)=9$ , and $z(9,0)=0$ , so at this point we get

$\dfrac{\partial w}{\partial u}(9,0)=2\cdot9\cdot\cos0=18$

7 0

3 years ago

Two numbers have a difference of 0.85 and a sum of 1 what are the numbers?

soldier1979 [14.2K]

X - y = 0.85
x + y = 1
----------------add
2x = 1.85
x = 1.85/2
x = 0.925

x + y = 1
0.925 + y = 1
y = 1 - 0.925
y = 0.075

so ur 2 numbers are : 0.925 and 0.075

3 0

3 years ago

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