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)
Answer : The two dimensions are the same in the triangles as they were in the rectangle.
- Rounding 789 nearest to tens will be 790.
Step-step Explanation:
Let's know how to round any number nearest to tens (Rules):-
We know tens place in any no. from left is at second place and number right to tens place need to be changed as:-
- If last digit is 0, then we don't have to do any rounding, because it is already to the ten.
- We round the no. down to the nearest ten if the last digit in the no. is 1, 2, 3, or 4
- We round the no. up to the nearest ten if the last digit in the no. is 5, 6, 7, 8, or 9.
Answer: 3/4
Step-by-step explanation:
Graph the line using the slope and y-intercept, or two points.
Slope: 3/4
HOPE THIS HELPS :)
