Answer and Explanation:
Given : Sides of right triangle 5,12 and 13.
To find : Show that the area of a right triangle of sides 5, 12 and 13 cannot be a square ?
Solution :
If 5,12 and 13 are sides of a right angle triangle then
13 is the hypotenuse as it is largest side.
then we take perpendicular as 12 and base as 5.
The area of the right angle triangle is
Here, h=12 and b=5
The area of the right angle triangle is 30 units.
30 is not a perfect square as 
There is no square pair formed.
Answer:
The vertex form is: y = 3(x-2)^2 + 7
. The vertex is (2, 7)
Step-by-step explanation:
Write the function: y = 3x^2 - 12x + 11 in vertex form
The vertex form of a quadratic equation is:
y = m(x - a)^2 + b where (a,b) is the vertex
For y = 3x^2 - 12x + 11
Solve for m
y = 3(x^2 - 4x) + 11
Complete the Square:
y = 3(x^2 - 4x + 4) + 11 - 4
y = 3(x - 2)^2 + 7
The vertex then is (2, 7)
119 times .5 equals $59.5
$119-$59.5=$59.5
They’re called transversal lines
Answer:
The coordinates of the point that makes the division in the given ratio is (0,3)
Step-by-step explanation:
Here, we want to find the point on the line segment that divides the line segment in the ratio 1:2
We simply use the internal division formula
That would be;
(x,y) = (nx1 + mx2)/(m + n) , (ny1 + my2)/(m + n)
m = 1 and n = 2
(x1,y1) = (4,9)
(x2,y2) = (-8,-9)
Substituting these values into the formula, we have;
2(4) + 1(-8)/(1 + 2) , 2(9) + 1(-9)/(2 + 1)
= (8-8)/3 , (18-9)/3
= (0,3)
