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3 years ago

PLEASE HELP I WILL GIVE 50 POINTS! IF YOU GIVE FAKE ANSWER I WILL REPORT YOU!!!

Mathematics

1 answer:

defon3 years ago

3 0

Answer:

It looks like your first account increases linearly,

unless you have a typo and meant:

f(x) = 10,125(1.83)^x

the second

exponential and can be written as

g(r) = 9,638(1.95)^r

clearly g(r) has a higher percentage change since 1.95 > 1.83

a more interesting question might have been: When will the amount be the same?

10,125(1.83)^x = 9,638(1.95)^x

(1.83)^x = .951901...(1.95)^x

take log of both sides and use log rules

xlog1.83 = log .951901... + log 1.95

x(log 1.83 - log 1.95) = log .951901...

x = .776..

check:

f(.776..) = 10,125(1.83)^.776.. = 1618410

g(.776..) = 9,638(1.95)^.776... = 16184.10

Step-by-step explanation:

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2 years ago

A class of 28 students has 12 girls. What is the ratio of girls to boys

DochEvi [55]

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3 girls for every 4 boys

Step-by-step explanation:

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3 years ago

The time, T (seconds) it takes for a pot of water to boil is inversely proportional to the cooker setting, H, applied to the pot

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3 years ago

How would you write a decimal that is less than five Ten thousandths

asambeis [7]

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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