Answer:
the sum of a four angles of quadrilateral is 180.<u>False</u>
<u>Note</u><u>:</u><u> </u><u>its</u><u> </u><u>3</u><u>6</u><u>0</u><u>°</u>
Answer:
The area of the square adjacent to the third side of the triangle is 11 units²
Step-by-step explanation:
We are given the area of two squares, one being 33 units² the other 44 units². A square is present with all sides being equal, and hence the length of the square present with an area of 33 units² say, should be x² = 33 - if x = the length of one side. Let's make it so that this side belongs to the side of the triangle, to our convenience,
x² = 33,
x =
.... this is the length of the square, but also a leg of the triangle. Let's calculate the length of the square present with an area of 44 units². This would also be the hypotenuse of the triangle.
x² = 44,
x =
.... applying pythagorean theorem we should receive the length of a side of the unknown square area. By taking this length to the power of two, we can calculate the square's area, and hence get our solution.
Let x = the length of the side of the unknown square's area -
=
+
,
x =
... And
squared is 11, making the area of this square 11 units².
Answer:
(x, y) = (-6, 0)
Step-by-step explanation:
The y-coefficients have opposite signs, so we can eliminate y-terms by multiplying both equations by a positive number and adding the results.
9 times the first equation plus 4 times the second gives ...
9(5x +4y) +4(3x -9y) = 9(-30) +4(-18)
45x +36y +12x -36y = -270 -72 . . . . eliminate parentheses
57x = -342 . . . . . collect terms
x = -6 . . . . . . . divide by the coefficient of x
__
Substituting into the first equation gives ...
5(-6) +4y = -30
4y = 0 . . . . . . . . . add 30
y = 0
The solution is (x, y) = (-6, 0).
Answer:
Perimeter = 22.809836694575
Step-by-step explanation:
A(-2, 2) ; B(6,2) ; C(0, 8)



Then
The perimeter of ΔABC = AB + BC + AC

= 22.809836694575