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expeople1 [14]
3 years ago
8

What are the coordinates of the circumcenter of a triangle with vertices A(−1, 1) , B(5, 1) , and C(−1, −1) ?

Mathematics
1 answer:
Elina [12.6K]3 years ago
7 0
Triangle ABC is a right triangle with vertex A containing the right angle.

You can tell because AB is a horizontal line since A and B have the same y-coordinate, and AC is a vertical line since A and C have the same x-coordinate.

A right triangle in a circumcircle is always in the semicircle, with the centre of the circle lying in the middle of the side opposite the right angle.

So, the circumcenter is the midpoint of BC.

The midpoint of BC is ((5-1)/2, (1-1)/2) = (2, 0)
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Shouldnt it be -8,4;7
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find the equation of a straight line passing through the point(4,5) and equally inclined to the lines 3x = 4y +7 and 5y= 12x +6
galina1969 [7]
<span>First we have to determine the slope of each lines by transforming to the slope-intercept form:

y=(3x-7/)4; m2= ¾y=(12x+6)/5, m3 = 12/5

The formula to be used in the proceeding steps is    a=tan^-1(m1-m2)/1+m1m2=tan^-1(m1-m2)/1+m1m2
substituting,      a=tan^-1(m1-3/4)/1+3m1/4=tan^-1(m1-12/5)1+12m1/5)     =>(4m1-3)/(4+3m1)=(5m1-12)/(5+12m1)m1 = -1applying this slope 

y -y1 = m(x-x1)
when y1 = 5 and x1 = 4 then,
y - 5 = -1(x-4)
y = -x +4+ 5 ; y = -x +9</span>
3 0
3 years ago
A number x rounded to 1 significant figure is 40. What is the error interval for x.
Viktor [21]

Answer:sorry i need points for my test

Step-by-step explanation:

5 0
3 years ago
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If 4x +5=9x-2,then x=
user100 [1]
<span>4x +5=9x-2
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5 0
3 years ago
H(x) = x2 1 k(x) = x – 2 (h k)(2) = (h – k)(3) = Evaluate 3h(2) 2k(3) =.
natima [27]

Quadratic equation is the equation in which only one variable is unknown. The highest power of the variable is 2.The value of the given functions are,

(h+k)(x)=5

(h-k)(x)=9

3h(2)+2k(3)=17

<h3>Given information-</h3>

The given function is,

h(x)=x^2+1

k(x)=x-2

<h3>Quadratic equation</h3>

Quadratic equation is the equation in which only one variable is unknown. The highest power of the variable is 2.

1) The value of the function (h+k)(2),

(h+k)(x)=h(x)+k(x)

(h+k)(x)=x^2+1+x-2

(h+k)(2)=2^2+1+2-2

(h+k)(x)=5

2)The value of the function (h-k)(3),

(h-k)(x)=h(x)-k(x)

(h-k)(x)=x^2+1-x+2

(h-k)(3)=3^2+1-3+2

(h-k)(x)=9

3) The value of the function 3h(2)+2k(3)

3h(x)+2k(x)=3x^2+3+2x-2\times 2

3h(2)+2k(3)=3\times2^2+3+2\times2-2\times 2

3h(2)+2k(3)=17

Hence the value of the given functions are,

(h+k)(x)=5

(h-k)(x)=9

3h(2)+2k(3)=17

Learn more about the quadratic equation here;

brainly.com/question/2263981

4 0
2 years ago
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