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Lina20 [59]
3 years ago
6

What are the first 20 digits in TT(pi)

Mathematics
2 answers:
bezimeni [28]3 years ago
5 0

Answer:

3,1,4,1,5,9,2,6,5,3,5,8,9,7,9,3,2,3,8,4

Step-by-step explanation:

3.14159265358979323846264338327950288419716

ad-work [718]3 years ago
3 0
Here are a couple.

<span>3.141592653589793238462643383279502884197169399375105
</span>
But it goes on and on.

I if got the answer wrong let me know.

Hope I helped.

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The answer is sixteen thousand four hundred ninety
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Gianna is preparing for the upcoming tennis season. According to her schedule, she can complete at most 10 workouts this week. I
Strike441 [17]
The whole number of minutes that Gianna complete each week as a function of x and y:
45x+30y
Since the above number is greater than 450 minutes, we get the inequality:
45x+30y\geq450
Also, "she can complete at most 10 workouts this week" so we deduce the inequality:
\\x+y\leq10
Therefore, the system of inequalities is the following:
45x+30y\geq450
\\x+y\leq10
The above system has many solutions, for example :
4 0
3 years ago
Rachel saved money to buy art supplies. She used 13 of her savings to buy brushes. She used 35 of her savings to buy paint. What
vekshin1

Given:

Rachel used \dfrac{1}{3} of her savings to buy brushes.

She used \dfrac{3}{5} of her savings to buy paint.

To find:

The fraction for remaining savings.

Solution:

Fraction that Rachel used of her savings to buy brushes and paint is

\text{Fraction for used amount}=\dfrac{1}{3}+\dfrac{3}{5}

                                    =\dfrac{5+9}{15}

                                    =\dfrac{14}{15}

Now,

\text{Fraction for remaining savings}=1-\text{Fraction for used amount}

\text{Fraction for remaining savings}=1-\dfrac{14}{15}

\text{Fraction for remaining savings}=\dfrac{1}{15}

Therefore, \dfrac{1}{15} of her savings is remaining.

6 0
3 years ago
Is 9 a possible solution for -2x + 4 ≥ -14?
konstantin123 [22]

Answer:

Yes

Step-by-step explanation:

-2(9)+4= -14  and -14 is equal to -14.

6 0
3 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
3 years ago
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