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Assoli18 [71]
3 years ago
5

If f(x)=4^ then which of the following is the value of f(-2)?

Mathematics
1 answer:
vredina [299]3 years ago
6 0

Answer:

Step-by-step explanation:

hello :

f(x)=4^x   so : f(-2)=4^(-2) = 1/4^2 = 1/16

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Which of the following is true?
Dahasolnce [82]

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the right ones are a c and d maybe

Step-by-step explanation:

these are probably wrong but I need my points so figure it out.

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Deon finished his math assignment in 2/5 hours. Then he completed his chemistry assignment in 1/2 hours. How much more time did
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Answer:

Deon spent 1/10 of an hour more on chemistry than on math

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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
I need help ASAP!!!!!!
topjm [15]

Answer:

1. YES  2. NO

Step-by-step explanation:

1. Since x=0.121212... and the question is asking for 100x, we need to multiply each side by 100. 0.121212... multiplied by 100 is 12.1212... So YES, 100x is still a repeating decimal.

2. 12.121212... minus 0.121212... would just be 12. This is because if we know that all the numbers behind the decimal point is 0.121212 and so on, 12.121212... minus 0.121212... the numbers behind the decimal would cancel out. So we are left with 12. So NO, the answer for 99x is NOT a repeating decimal.

6 0
3 years ago
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