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MissTica
2 years ago
12

On his first day working out, anthony did 30 push-ups. on each successive day, he did exactly 3 more push-ups than on the previo

us day. after completing his push-ups on the 30th day, how many push-ups had he completed on all 30 days?
Mathematics
2 answers:
babymother [125]2 years ago
5 0
The correct answer would be more than 2,000 (approximately 2,205) because on the first he did 30, the second was 33, third was 36, and so on. You would have to make a chart for this, which could be somewhat boring, or you could write it out and do 30+30+3+30+6+30+9, etc for 30 different days, with 30 being for each day and progressively adding 3. Hope this helped!
Lelechka [254]2 years ago
3 0
(a₁)=30; n=30; d=3; S₃₀=(n/2)*(2a₁+(n-1)*d)=15*(60+29*3)=2205
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2 years ago
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
2 years ago
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