Answer:
26.2 units
Step-by-step explanation:
We are given the points/vertices
A(6, 3),
B(6, - 2) , and
C(- 4, 3)
Step two
Let us find the distances between the given points/vertices
A-B =A(6, 3) to B(6,-2)
d=√((x2-x1)²+(y2-y1)²)
Substitute
d=√((6-6)²+(-2-3)²)
d=√(-2-3)²)
d=√(-5)²)
d=5 units
B-C=B(6, - 2) to C(-4, 3)
d=√((x2-x1)²+(y2-y1)²)
Substitute
d=√((-4-6)²+(3+2)²)
d=√(-10)²+(5)²)
d=√100+25
d=√125
d=11.2 units
C-A=C(-4, 3) to A(6, 3)
d=√((x2-x1)²+(y2-y1)²)
Substitute
d=√((6+4)²+(3-3)²)
d=√(10)²
d=√100
d=10 units
Hence the perimeter is 5+11.2+10
P=26.2 units
Step-by-step explanation:
it is the ans to the midpoint.
We know that
the distance from the centroid of the triangle to one of the vertices is the radius of the circle <span>required to inscribe an equilateral triangle.
[distance </span>centroid of the triangle to one of the vertices]=(2/3)*h
h=the <span>altitude of the equilateral triangle-----> 5.196 in
so
</span>[distance centroid of the triangle to one of the vertices]=(2/3)*5.196
[distance centroid of the triangle to one of the vertices]=3.464 in----> 3.5 in
the radius is equal to the distance of the centroid of the triangle to one of the vertices
hence
the radius is 3.5 in
the answer is
the radius is 3.5 in
Answer:
x = 1
, y = 3 thus: A is your Anser
Step-by-step explanation:
Solve the following system:
{2 x + y = 5 | (equation 1)
x + y = 4 | (equation 2)
Subtract 1/2 × (equation 1) from equation 2:
{2 x + y = 5 | (equation 1)
0 x+y/2 = 3/2 | (equation 2)
Multiply equation 2 by 2:
{2 x + y = 5 | (equation 1)
0 x+y = 3 | (equation 2)
Subtract equation 2 from equation 1:
{2 x+0 y = 2 | (equation 1)
0 x+y = 3 | (equation 2)
Divide equation 1 by 2:
{x+0 y = 1 | (equation 1)
0 x+y = 3 | (equation 2)
Collect results:
Answer: {x = 1
, y = 3
Answer:
The probability that a student plays either basketball or soccer is 19% or 0.19.
Step-by-step explanation:
Let A be the event that student play basketball and B be the event that student play soccer.
It is given that 6 students play on both teams.
We have to find the probability that a student plays either basketball or soccer.
Therefore the probability that a student plays either basketball or soccer is 19% or 0.19.
