Answer:
2
Step-by-step explanation:
when it returns to the ground , h = 0
32t -16t2 = 0 and solve for t
t = 0, 2 ....0 is the start, 2 is when it returns
Answer:
Step-by-step explanation:
In Δ ABC & ΔADC
AC bisect ∠BCD given
∠ACB ≅ ∠ACB AC bisect ∠BCD
DC ⊥ AD Given
BC ⊥ AB Given
∠ABC = ∠ADC Right angles are ≅
AC ≅ AC Common to both triangle ΔABC & ΔADC
ΔABC ≅ ΔADC AAS congruent
BC ≅ DC CPCT
Answer:
21.38
Step-by-step explanation:
Find the diagram attached
In order to get PS, we need to get the length of QS first
According to triangle QRS;
QS =opposite
QR = 39(adjacent)
Using SOH CAH TOA
Tan theta = opp/adj
Tan 28 = QS/39
QS = 39tan28
QS = 39(0.5317)
QS = 20.74
Get PS
sin theta = QS/PS
sin 76 = 20.74/PS
PS = 20.74/sin76
PS = 20.74/0.9703
PS = 21.38
Hence the length of PS is 21.38
The complete question is
Find the volume of each sphere for the given radius. <span>Round to the nearest tenth
we know that
[volume of a sphere]=(4/3)*pi*r</span>³
case 1) r=40 mm
[volume of a sphere]=(4/3)*pi*40³------> 267946.66 mm³-----> 267946.7 mm³
case 2) r=22 in
[volume of a sphere]=(4/3)*pi*22³------> 44579.63 in³----> 44579.6 in³
case 3) r=7 cm
[volume of a sphere]=(4/3)*pi*7³------> 1436.03 cm³----> 1436 cm³
case 4) r=34 mm
[volume of a sphere]=(4/3)*pi*34³------> 164552.74 mm³----> 164552.7 mm³
case 5) r=48 mm
[volume of a sphere]=(4/3)*pi*48³------> 463011.83 mm³----> 463011.8 mm³
case 6) r=9 in
[volume of a sphere]=(4/3)*pi*9³------> 3052.08 in³----> 3052 in³
case 7) r=6.7 ft
[volume of a sphere]=(4/3)*pi*6.7³------> 1259.19 ft³-----> 1259.2 ft³
case 8) r=12 mm
[volume of a sphere]=(4/3)*pi*12³------>7234.56 mm³-----> 7234.6 mm³
