Answer:
(6,1)
Step-by-step explanation:
If you plug these numbers in for x and y, the equation works.
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:56.8 inches
Step-by-step explanation:
For perimeter we must find the lengths of the 2 diagonal sides of the triangle which are the same because it is isosceles.
the triangle is cut directly in half so both halves of the base are 17/2
17/2 = 8.5
solve through pythagoren theorum
altitude^2 + (half of base)^2 = (diagonal length)^2
18^2 + (8.5)^2 = (diagonal length)^2
324 + 72.25 = (diagonal length)^2
396.3 = (diagonal length)^2
= diagonal length
diagonal length = 19.9
diagonal length + diagonal length + base = perimeter
19.9 + 19.9 + 17 = perimeter
56.8 inches = perimeter
Answer:
25%
Step-by-step explanation:
An easy way to figure this out is just by dividing the initial amount of students (18) by the current amount of students (24), and then subtracting 1.
So, 18/24 = 0.75 (aka 75%), and 1 - 0.75 = 0.25 (aka 25%)
Answer:
27
Step-by-step explanation:
First you need to find x.
It is shown that DE=EF, so we can use this.
5x-3=3x+7
5x=3x+10
2x=10
x=5
Then, you plug in the value of the variable and solve.
6(5)-3
30-3
27 = DF
