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inysia [295]
3 years ago
8

Are the statements true or false?

Mathematics
1 answer:
MrMuchimi3 years ago
3 0

Answer:

It Is true and then false

Step-by-step explanation:

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PLEASE HELP ASAP!!!!
Paul [167]

Answer:

the answers are <u>B and D </u>

Step-by-step explanation:

please let me know if I am wrong

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3 years ago
How many 1/2 pound bags can fill 14 pounds
faltersainse [42]
It’s easy do 2 times 14 so it’s 28
4 0
3 years ago
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Find the area for the following figure.<br> O 124. 86 cm2<br> 62.93 cm2<br> 26.5 cm2<br> 137.45 cm2
Semmy [17]
The answer is 62.93
8 0
3 years ago
If a car with an average speed of 40 miles per hour drives for 45 minutes how far will it have gone
REY [17]

Answer: 30\ miles

Step-by-step explanation:

The average speed of the car is 40 miles per hour:

V=40\ \frac{mi}{h}

The car drives for 45 minutes:

t=45\ min

Since 1 hours has 60 minutes, we can convert the time from minutes to hours:

(45\ min)(\frac{1\ h}{60\ min})=0.75\ h

Knowing these values, we can use the following formula for calculate the distace "d":

d=V*t

Therefore, substituting values into the formula, we get:

d=(40\ \frac{mi}{h})(0.75\ h)\\\\d=30\ mi

3 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
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