Answer:
A
Step-by-step explanation:
● first one:
The diagonals of a rhombus are perpendicular to each others wich means that they form four right angles.
STP is one of them so this statement is true.
● second one:
If ST and PT were equal this would be a square not a rhombus.
● third one:
If SPQ was a right angle, this woukd be a square.
● fourth one:
Again if the diagonals SQ and PR were equal, this would be a square.
Answer:
16 - 5 
Step-by-step explanation:
Answer:
Step-by-step explanation:
The solution is in the attached file
Find the possible rational roots and use synthetic division to find the first zero.
I chose x=1 (which represents the factor "x-1")
1║2 -7 -13 63 -45
║ 2 -5 -18 45
2 -5 -18 45 0
(x-1) is a factor, (2x³ - 5x² - 18x + 45) is the other factor.
Use synthetic division on the decomposed polynomial to find the next zero.
I chose x = 3 (which represents the factor "x-3")
3║2 -5 -18 45
║ 6 3 -45
2 1 -15 0
Using synthetic division, we discovered that (x-1), (x-3), & (2x² + x -15) are factors. Take the new decomposed polynomial (2x² + x -15) and find the last two factors using any method.
Final Answer: (x-1)(x-3)(x+3)(2x-5)
Check the picture below.
let's recall that a kite is a quadrilateral, and thus is a polygon with 4 sides
sum of all interior angles in a polygon
180(n - 2) n = number of sides
so for a quadrilateral that'd be 180( 4 - 2 ) = 360, thus
![\bf 3b+70+50+3b=360\implies 6b+120=360\implies 6b=240 \\\\\\ b=\cfrac{240}{6}\implies b=40 \\\\[-0.35em] ~\dotfill\\\\ \overline{XY}=\overline{YZ}\implies 3a-5=a+11\implies 2a-5=11 \\\\\\ 2a=16\implies a=\cfrac{16}{2}\implies a=8](https://tex.z-dn.net/?f=%5Cbf%203b%2B70%2B50%2B3b%3D360%5Cimplies%206b%2B120%3D360%5Cimplies%206b%3D240%20%5C%5C%5C%5C%5C%5C%20b%3D%5Ccfrac%7B240%7D%7B6%7D%5Cimplies%20b%3D40%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Coverline%7BXY%7D%3D%5Coverline%7BYZ%7D%5Cimplies%203a-5%3Da%2B11%5Cimplies%202a-5%3D11%20%5C%5C%5C%5C%5C%5C%202a%3D16%5Cimplies%20a%3D%5Ccfrac%7B16%7D%7B2%7D%5Cimplies%20a%3D8)
