X(7x+3)=0
x=0
or
7x=-3
x=-3/7
so,
(0,0) or (-3/7,0)
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
(A) For x representing the cost of one of Tanya's items, her total purchase cost 5x. The cost of one of Tony's items is then (x-1.75) and the total of Tony's purchase is 6(x-1.75). The problem statement tells us these are equal values. Your equation is ...
... 5x = 6(x -1.75)
(B) Subtract 5x, simplify and add the opposite of the constant.
... 5x -5x = 6x -6·1.75 -5x
... 0 = x -10.50
... 10.50 = x
(C) 5x = 5·10.50 = 52.50
... 6(x -1.75) = 6·8.75 = 52.50 . . . . . the two purchases are the same value
(D) The individual cost of Tanya's iterms was $10.50. The individual cost of Tony's items was $8.75.
Answer:
0.800
Step-by-step explanation:
Answer:
The teacher is 1.416667 times taller than the student
Step-by-step explanation:
(5 2/3)/4
=1.416667