3/3...............................
Answer:
Time t = 2 seconds
It will reach the maximum height after 2 seconds
Completed question;
Amir stands on a balcony and throws a ball to his dog who is at ground level. The ball's height, in meters above the ground, after t seconds that Amir has thrown the ball is given by:
H (t) = -(t-2)^2+9
many seconds after being thrown will the ball reach its maximum height?
Step-by-step explanation:
The equation of the height!
h(t) = -(t-2)^2 + 9 = -(t^2 -4t +4) + 9
h(t) = -t^2 +4t -4+9
h(t) = -t^2 + 4t +5
The maximum height is at dh/dt = 0
dh/dt = -2t +4 = 0
2t = 4
t = 4/2 = 2
Time t = 2 seconds
It will reach the maximum height after 2 seconds
Answer:
1. Yes
∆RST ~ ∆WSX
by SAS
2. Yes
∆ABC ~ ∆PQR
by SSS
3. Yes
∆STU ~ ∆JPM
by SAS
4. Yes
∆DJK ~ ∆PZR
by SAS
5. Yes
∆RTU ~ ∆STL
by SAS
5. Yes
∆JKL ~ ∆XYW
by SAS
6. No
7. Yes
∆BEF ~ ∆NML
by SAS
8. Yes
∆GHI ~ ∆QRS
by SSS
9. x=22
10. x=12
Step-by-step explanation:
1. RS/WS=ST/SX and m<RST=m<WSX
2. AB/PQ=8/6=4/3
BC/QR=AC/PR=12/9=4/3
AB/PQ=BC/QR=AC/PR
3. ST/JP=10/15=2/3
SU/JM=14/21=2/3
ST/JP=2/3=SU/JM
and m<TSU=70°=m<PJM
4. DK/PR=8/4=2
JK/ZR=18/9=2
DK/PR=2=JK/ZR
and m<DKJ=65°=m<PRZ
5. RT/ST=UT/LT
and m<RTU=m<STL
6. KL/YW=20/18=10/9
JL/XW=36/24=3/2
KL/YW=10/9≠3/2=JL/XW
7. BF/NL=24/16=3/2
BE/NM=39/26=3/2
BF/NL=3/2=BE/NM
and m<EBF=m<MNL
8. GH/QR=32/20=8/5
HI/RS=40/25=8/5
GI/QS=24/15=8/5
GH/QR=HI/RS=GI/QS=8/5
9. x/33=18/27
Simplifying the fraction on the right side of the equation:
x/33=2/3
Solving for x: Multiplying both sides of the equation by 33:
33(x/33)=33(2/3)
x=11(2)
x=22
10. x/16=9/12
Simplifying the fraction on the right side of the equation:
x/16=3/4
Solving for x: Multiplying both sides of the equation by 16:
16(x/16)=16(3/4)
x=4(3)
x=12
Answer:
96
Step-by-step explanation:
commom difference = d
a₆ + a₇ = 16
a₅ + a₈ = (a₆ - d) + (a₇ + d) = a₆ + a₇ = 16
a₄ + a₉ = (a₆ - 2d) + (a₇ + 2d) = a₆ + a₇ = 16
Similarly,
a₃ + a₁₀ = 16, a₂ + a₁₁ = 16, a₁ + a₁₂ = 16
so
a₁ + a₂ + ... + a₁₁ + a₁₂ = 6 x 16 = 96
