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
I don't think this is middle school...
(2^8 x 3^-5 x 6^0)^-2 x ((3^-2)/(2^3))^4 x (2^28)
(256 x (1/243))^-2 x ((1/9)/(8))^4 x 268435456
1.05349794239^-2 x (<span>0.01388888888)^4 x 268435456
</span><span>
0.90101623534 x </span><span>3.72108862e-8 x 268435456
</span>8.99999997623
9
9 is your answer.
Answer:
critical value = 5.29
Step-by-step explanation:
Given that they are divided into 4 groups and a sample of 5 test was selected
N = 5 * 4 = 20
k = 4
∝ = 0.01
Df for numerator ( SS group )= k - 1 = 3
Df for denominator ( SSE group ) = N - k = 20 - 4 = 16
DF ( degree of freedom )
Next we will use the F table to determine the critical value
Critical value = 
= 5.29
Answer:
d(x) = 3.5x + 0.5
Step-by-step explanation:
3.5 x 5 = 17.5 which means the canoe began from a starting point of 0.5
Answer:
x = 20
Step-by-step explanation:
hmu for more information
