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Ganezh [65]
3 years ago
14

Complete the two column proof Given: 22= 24,m_2 = 110°

Mathematics
1 answer:
Savatey [412]3 years ago
6 0

Step-by-step explanation:

I took 3 indicated as 5 and its adjacent angle to be 3

<2 = <4

As <2 =<3( corresponding angle)

And <3 = <4 ( Vertically opp.angle)

hence <2 = <4

<2 =>110

so , <4 = 110

So, <4 =>110

As <4 and <5 form linear pair

So <4 + <5 =>180

<5 = 180 -110 =>70

As i took <5 as replacing angle to <3

So According to Question fig

<3 =>70

Hence proved

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Write fraction in "Simpilist Form". If already , then write "Simpilist Form". The Fraction is: 3 over 100 - 3/100​
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see explanation

Step-by-step explanation:

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Nina and Jo both ran an 8 km race. Nina took 55 minutes to run the whole race. Jo started the race 3 minutes later than Nina but
grin007 [14]

Answer:

Step-by-step explanation:

First we figure out how fast Nina can run. If Nina can run 8 km in 55 minutes, then her rate is

\frac{8km}{55min}=.145\frac{km}{min} and we can use that in a d = rt table:

                 d        =        r        *        t

Nina                            .145

Jo

Now we can fill in the distance which is 6 for both, since that is the distance where they met:

                d        =        r       *        t

Nina         6        =     .145

Jo             6        =

Now we go to the info given about the time. If Jo started the race 3 minutes after Nina, that means that Nina is running 3 minutes longer than Jo. Filling in the time info:

                d        =        r        *        t

Nina          6       =       .145    *      t + 3

Jo              6       =         r       *         t

As you can see, right now we have 2 unknowns in Jo's row. But we don't have to! We will go to Nina's row where the only unknown is time and solve for t. If d = rt, then

6 = .145(t + 3) and

6 = .145t + .435 and

5.55 = .145t so

t = 38.379 minutes. This means that Jo was running 38.379 minutes when she caught up to Nina (it took Nina 3 minutes longer than that to go 6 km since she was already running for 3 minutes when Jo started the race). If Jo's time is 38.379, we can use that in her row for t and solve for r. If d = rt, then

6 = r(38.379) and

r = .16 km/min

Let's check it without the rounding (rounding takes away from the accuracy). If 6 = .145(t + 3) and Nina's rate not rounded is .145454545 and t = 38.37931034, then, rewriting without rounding:

6 should equal .145454545( 38.37931034 + 3)

6 ?=? .145454545(41.37931034)

6 ?=? 6.0 so

Jo's rate is .16 km/min rounded

6 0
3 years ago
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