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USPshnik [31]
3 years ago
7

pam's weekly exercise schedule is shown in the bar graph below. on what day did pam exercise for 45 minutes ? how long did pam e

xercise on thursday ? how many days did pam exercise that week ?

Mathematics
1 answer:
gavmur [86]3 years ago
5 0
She exercised 45 minutes on friday.
she exercised 30 minutes on thursday
she exercised 5 days that week
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Adult - 2 tickets . child- 9 tickets.
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2 years ago
What’s the correct answer for this?
Alexxx [7]

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the answer is the top choice

3 0
3 years ago
A triangular courtyard has a perimeter of 120 meters. The lengths of two sides are 30 meters and 50 meters. How long is the thir
Masja [62]

Answer:

the third side is 40 m long

Step-by-step explanation:

The perimeter of the triangle is the sum of its three sides, and they give you what that value in meters is (120 m)

Your are given the length of two of them: 30 m and 50 m, and need to find the third one (let's call it "x" for this unknown side)

Now set the following equation:

Perimeter = side 1 + side 2 + side 3   --> replace these with the info you know

120 m = 30 m + 50 m + x   --> add 30 m and 50 m obtaining 80 m

120 m = 80 m + x  --> now solve for x (isolate the x on one side) by subtracting 80 m from both sides

120 m - 80 m = x  --> perform the subtraction 120 m - 80 m = 40 m

40 m = x

Which tells us that the third unknown side has a length of 40 m

7 0
3 years ago
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Lelechka [254]

Answer:

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3 0
2 years ago
If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.
Pie
Given that f^{(n)}(0)=(n+1)!, we have for f(x) the Taylor series expansion about 0 as

f(x)=\displaystyle\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n

Replace n+1 with n, so that the series is equivalent to

f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}

and notice that

\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}

Recall that for |x|, we have

\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}

which means

f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}
\implies f(x)=\dfrac1{(1-x)^2}
5 0
3 years ago
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