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LiRa [457]
4 years ago
14

63 people are going to the zoo. There are 3 cars to take people to the zoo. How

Mathematics
1 answer:
fomenos4 years ago
4 0

Answer:

21

3x21= 63 have a great day

Step-by-step explanation:

BRAINLIEST PLEASE

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3 years ago
1.The mass of an object, usually measured in kilograms and grams is?
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Answer:

1.kg(not sure)

2. mass

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frank made $156.00 12 hours of work. at the same rate, how many hours would he have to work to make $221?
uranmaximum [27]

Answer:

17 hours

Step-by-step explanation:

We first need to find the hourly rate:

156 / 12 = 13

Hourly Rate: $13

Now we can divide 221 by 13, this gives us 17.

That means that Frank will have to work 17 hours to make $221.

Hope this helps!

7 0
3 years ago
Que is on pic.i can't able to type in text.
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It's not difficult to compute the values of A and B directly:

A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}
A=\tan^{-1}(\sin\theta)-\dfrac\pi4

B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt
B=\left(\ln|t|-\dfrac12\ln|1+t^2|\right)\bigg|_{t=1}^{t=\csc\theta}
B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2

Let's assume 0, so that |\csc\theta|=\csc\theta.

Now,

\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}
\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^{A+B}\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}
\Delta=A(-B^2+A^2+B^2)-e^{A+B}(-A^2-A^2B-B^3)+(-A^2-B^3)
\Delta=A^3-A^2-B^3+e^{A+B}(A^2+A^2B+B^3)

There doesn't seem to be anything interesting about this result... But all that's left to do is plug in A and B.
3 0
3 years ago
John runs to the market and comes back in 15 minutes. His speed on the way to the market is 5m/s and his speed on the way back i
katen-ka-za [31]

Answer:

The distance to the market is 2000 m

Step-by-step explanation:

∵ John runs to the market and comes back in 15 minutes

→ Change the min. to the sec. because the unit of his speed is m/s

∵ 1 minute = 60 seconds

∴ 15 minutes = 15 × 60 = 900 seconds

→ Assume that t1 is his time to the market and t2 is his time from

  the market

∵  t1 + t2 = 15 minutes

∴ t1 + t2 = 900 ⇒ (1)

→ Assume that the distance to the market is d

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∵ Time = Distance ÷ Speed

∴ t1 = d ÷ 5 ⇒ (1 ÷ 5 = 0.2)

∴ t1 = 0.2d ⇒ (2)

∵ His speed on the way back is 4m/s

∴ t2 = d ÷ 4 ⇒ (1 ÷ 4 = 0.25)

∴ t2 = 0.25d ⇒ (3)

→ Substitute (2) and (3) in (1)

∵ 0.2d + 0.25d = 900

∴ 0.45d = 900

→ Divide both sides by 0.45

∴ d = 2000 m

∴ The distance to the market = 2000 m

8 0
3 years ago
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