First, 9.8 (Gravity) times 3 (time) equals 29.4, which is the velocity after 3 seconds. The kinematic equation for change in position that uses the variables we have is:
delta x= (v)(t) -0.5(acceleration)(time)^2
delta x= 29.4 times (3) - 0.5 (9.8) times 9
delta x= 44.1
100 minus 44.1 equals 55.9, which is the answer for part a.
Tell me if you need any clarification
PART B:
The kinematic equation for this is:
delta x= (initial velocity) times time plus 0.5 (a)(time)^2
100=(0)times(x) plus 0.5 (a)(time)^2
100=0.5(9.8)(x)^2
100=4.9x^2
100/4.9 is approxamitely 20.4.
The squareroot of this is approxamitely 4.5.
4.5 seconds
Tell me if you need any clarification
Answer:
√2=1.414
then :√8 +2√32 +3√128+4√50
√8=√2³ =2√2
2√32=√2^5 = 4*2√2 = 8√2
3√128 = 3√2^6*2=8*3√2 =24√2
4√50 =4√5²*2= 20√2
add results : 2√2+8√2 +24√2+20√2=54√2
<h2>
54√2=54×1.414=76.356 ( it is not in the options)</h2>
x=7-4√3
√x+ 1/√x
√(7-4√3) +1/√(7-4√3) =
(8-4√3)/√(7-4√3)
(8-6.93)/√(7-6.93) = 4 ( after rounded to the nearest whole number)
4 is your answer
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:
18
Step-by-step explanation:
