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Paha777 [63]
3 years ago
12

I need it fasttttttt

Mathematics
2 answers:
Rzqust [24]3 years ago
4 0

Answer:

well, there is already $900 so 1800-900=900   now we need 4% of 900 which is $36   900 divided by 36 is 25

it will take 25 years

Step-by-step explanation:

plz mark brainliest i answered first

Olenka [21]3 years ago
3 0

the answer to this question is 15

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A function and its inverse are reflected over the line y = x. To graph a function and its inverse, all you have to do is graph the function and then switch all x and y values in each point to graph the inverse.
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Help me out please i will give branliest and 25 points !!!
Margarita [4]

Answer:

Answers A (1/3), C (4/15) and D (5/6) are all repeating decimals.  

Step-by-step explanation:

Repeating decimals are decimals in which a number or sequence of numbers is repeated over and over when dividing the numerator by the denominator of a fraction.  For example, when you divide 1 by 3 (1/3), you will get a repeating decimal of 0.3333...since 3 goes into 10 three times with a remainder of 1 and will keep going.  Likewise, when you divide 4/15, you get an initial value of 0.2, with a repeating 6, or 0.266666...  Lastly, when you divide 5/6, you will get a repeating decimal of 0.833333....  The other answers will all be terminating decimals when you divide the numerator by the denominator.  Terminating decimals mean the stop at some point and don't continue.  

6 0
3 years ago
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Y = |x-4|+6 The axis of symmetry is x=
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Step-by-step explanation:

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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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