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zysi [14]
3 years ago
9

PLZ HELP William’s typical running speed is 6 km/h. The difference between his fastest speed and his typical speed is 1.5 km/h.

The equation f-1.5=6 can be used to represent this problem, where f is William’s fastest speed. What is William’s fastest speed?
Mathematics
2 answers:
Jet001 [13]3 years ago
7 0

Answer:

William's fastest speed is 7.5 km/h

Step-by-step explanation:

Let the fastest speed of William is f therefore from the question's statement the equation will be

f - 6 = 1.5

Here difference between William's fastest speed and typical speed is 1.5 km/h

Now we will add 6 on both sides of the equation

(f - 6) + 6 = 1.5 + 6

f = 7.5 km/h

So the fastest speed is 7.5 km/h

yarga [219]3 years ago
4 0
I might not be right but I think William's fastest speed is 7.5 km/h. 
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olga_2 [115]

Answer:

1. Yes

∆RST ~ ∆WSX

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2. Yes

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3. Yes

∆STU ~ ∆JPM

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4. Yes

∆DJK ~ ∆PZR

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5. Yes

∆RTU ~ ∆STL

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5. Yes

∆JKL ~ ∆XYW

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6. No

7. Yes

∆BEF ~ ∆NML

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8. Yes

∆GHI ~ ∆QRS

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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

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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

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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

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3 years ago
In a newspaper, it was reported that yearly robberies in Springfield were up 26% to 252 in 2011 from 2010. How many robberies we
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Answer:

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Step-by-step explanation:

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4 years ago
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