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Vesnalui [34]
3 years ago
11

How many 1/3 are there in 32/3?

Mathematics
2 answers:
STALIN [3.7K]3 years ago
6 0
Answer: 32
1/3 * 32 = 32/3
Afina-wow [57]3 years ago
4 0
You'd need to divide

(32/3)/(1/3)
=(32/3)(3/1)
=96/3
=32
There are therefore 32
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Fifty balls numbered 1 through 50 are mixed up in a barrel. How many balls must you draw from the barrel (without looking) to be
ipn [44]
If you draw only 25 balls, you could draw the 25 odd-numbered balls.  However, there will then be no odd-numbered balls left, so when you draw two more you will be guaranteed to get two even-numbered balls.  Thus, the minimum is 25+2=27 balls.
8 0
3 years ago
On this corrdinate plane UVW has been transformed to form image U’ V’ W’.
egoroff_w [7]

Answer:

C

Step-by-step explanation:

Translate the image 2 units right and 1 unit up. Then rotate the image 180°.

Take the coordinate W, it is at (2, 4).

Translate 2 units right (add 2 to the x coordinate) and 1 up (add 1 to the y coordinate)

(2, 4) ------> (2 + 2, 4 + 1) -------> (4, 5)

A rotation of 180° (doesn't matter the direction) makes the coordinates their opposites. Positives become negatives and negatives become positive.

(4, 5) -------> (-4, -5)

8 0
2 years ago
Each week, Stephanie is paid a $180 base salary and a 3% commission on her total sales generated for the week. Which expression
ivanzaharov [21]

Answer:d

Step-by-step explanation:

Brainliest plz

8 0
3 years ago
Find the smallest 4 digit number such that when divided by 35, 42 or 63 remainder is always 5
alex41 [277]

The smallest such number is 1055.

We want to find x such that

\begin{cases}x\equiv5\pmod{35}\\x\equiv5\pmod{42}\\x\equiv5\pmod{63}\end{cases}

The moduli are not coprime, so we expand the system as follows in preparation for using the Chinese remainder theorem.

x\equiv5\pmod{35}\implies\begin{cases}x\equiv5\equiv0\pmod5\\x\equiv5\pmod7\end{cases}

x\equiv5\pmod{42}\implies\begin{cases}x\equiv5\equiv1\pmod2\\x\equiv5\equiv2\pmod3\\x\equiv5\pmod7\end{cases}

x\equiv5\pmod{63}\implies\begin{cases}x\equiv5\equiv2\pmod 3\\x\equiv5\pmod7\end{cases}

Taking everything together, we end up with the system

\begin{cases}x\equiv1\pmod2\\x\equiv2\pmod3\\x\equiv0\pmod5\\x\equiv5\pmod7\end{cases}

Now the moduli are coprime and we can apply the CRT.

We start with

x=3\cdot5\cdot7+2\cdot5\cdot7+2\cdot3\cdot7+2\cdot3\cdot5

Then taken modulo 2, 3, 5, and 7, all but the first, second, third, or last (respectively) terms will vanish.

Taken modulo 2, we end up with

x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod2

which means the first term is fine and doesn't require adjustment.

Taken modulo 3, we have

x\equiv2\cdot5\cdot7\equiv70\equiv1\pmod3

We want a remainder of 2, so we just need to multiply the second term by 2.

Taken modulo 5, we have

x\equiv2\cdot3\cdot7\equiv42\equiv2\pmod5

We want a remainder of 0, so we can just multiply this term by 0.

Taken modulo 7, we have

x\equiv2\cdot3\cdot5\equiv30\equiv2\pmod7

We want a remainder of 5, so we multiply by the inverse of 2 modulo 7, then by 5. Since 2\cdot4\equiv8\equiv1\pmod7, the inverse of 2 is 4.

So, we have to adjust x to

x=3\cdot5\cdot7+2^2\cdot5\cdot7+0+2^3\cdot3\cdot5^2=845

and from the CRT we find

x\equiv845\pmod2\cdot3\cdot5\cdot7\implies x\equiv5\pmod{210}

so that the general solution x=210n+5 for all integers n.

We want a 4 digit solution, so we want

210n+5\ge1000\implies210n\ge995\implies n\ge\dfrac{995}{210}\approx4.7\implies n=5

which gives x=210\cdot5+5=1055.

5 0
3 years ago
There is one and only one EQUATION for every Line in the coordinate plane.<br> O True<br> O False
yarga [219]

Answer:

False

Step-by-step explanation:

6 0
3 years ago
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