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saul85 [17]
3 years ago
13

When a teacher divided her students into groups of four, she had three students remaining. when she divided them into groups of

five, she had four students remaining. there were fewer than 40 students in the class. how many students could be in the class?
Mathematics
1 answer:
AfilCa [17]3 years ago
5 0
39 students....39/4 = 9 remainder 3
                       39/5 = 7 remainder 4

ur answer is 39
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cricket20 [7]
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4 0
1 year ago
The GCF of 24 and 32 is
Oksana_A [137]

Start by setting up a factor tree for each number.

Factor in anyway that you'd like.

At the end, circle all prime numbers.

Remember, prime numbers are numbers

with only 2 factors, 1 and the number itself.

Now write these factors in least to greatest

multiplying them separate rows.

Circle factors they have in common.

Since 3 pairs of two's match up, multiply by 2 three times to get 8.

So the GCF is 8.

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3 years ago
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2 years ago
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Elina [12.6K]

You want to end up with A\sin(\omega t+\phi). Expand this using the angle sum identity for sine:

A\sin(\omega t+\phi)=A\sin(\omega t)\cos\phi+A\cos(\omega t)\sin\phi

We want this to line up with 2\sin(4\pi t)+5\cos(4\pi t). Right away, we know \omega=4\pi.

We also need to have

\begin{cases}A\cos\phi=2\\A\sin\phi=5\end{cases}

Recall that \sin^2x+\cos^2x=1 for all x; this means

(A\cos\phi)^2+(A\sin\phi)^2=2^2+5^2\implies A^2=29\implies A=\sqrt{29}

Then

\begin{cases}\cos\phi=\frac2{\sqrt{29}}\\\sin\phi=\frac5{\sqrt{29}}\end{cases}\implies\tan\phi=\dfrac{\sin\phi}{\cos\phi}=\dfrac52\implies\phi=\tan^{-1}\left(\dfrac52\right)

So we end up with

2\sin(4\pi t)+5\cos(4\pi t)=\sqrt{29}\sin\left(4\pi t+\tan^{-1}\left(\dfrac52\right)\right)

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seraphim [82]
The problem seems confusing is it supposed to say y=?
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3 years ago
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