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erastovalidia [21]
3 years ago
13

Should 0 be on your number line? Explain.

Mathematics
2 answers:
artcher [175]3 years ago
5 0

Answer:

Yes

Step-by-step explanation:

That is where you start from. Before 1 is 0, so I would put 0 on your number line. Also, number lines have 0s on them. I don't think I ever saw a number line that doesn't have 0.

Hope I can help.

Vladimir [108]3 years ago
4 0

Yes it should, 0 is still a number even though it’s value is nothing. You can’t get to -1 without 0.

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agasfer [191]
I believe it would be $6
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3 years ago
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A certain country has $10 billion in paper currency in circulation, and each day $50 million comes into the country's banks. The
mash [69]

Answer:

\bf x(t)=10(1-e^{-0.005*t})

Step-by-step explanation:

The differential equation

\bf \displaystyle\frac{dx}{dt}=0.005(10-x)

can be solved by separation of variables. Write the equation as

\bf \displaystyle\frac{dx}{10-x}=0.005dt

Integrate on both sides

\bf \int\displaystyle\frac{dx}{10-x}=\int0.005dt\Rightarrow -ln(10-x)=0.005t+C\Rightarrow\\\\\Rightarrow ln(10-x)^{-1}=0.005t+C\Rightarrow (10-x)^{-1}=e^{0.005t}e^C

where C is a constant.

\bf e^C is also a constant and we will keep calling it C, (there is no reason to change the letter). We have then

\bf (10-x)^{-1}=Ce^{0.005t}\Rightarrow \displaystyle\frac{1}{10-x}=Ce^{0.005t}\Rightarrow 10-x=\displaystyle\frac{1}{Ce^{0.005t}}\Rightarrow\\\\\Rightarrow x(t)=10-(1/C)e{-0.005t}

(1/C) is a constant, and for the same reason we will keep calling it C. So the general solution is

\bf x(t)=10-Ce^{-0.005t}

Now, we use the initial condition x(0)=0

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and the particular solution is

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7 0
3 years ago
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Alik [6]

Answer:

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Step-by-step explanation:

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8 0
3 years ago
What is the factored form of the polynomial? z2 − 10z + 25
Crank

Answer:

(z−5)(z−5)

Step-by-step explanation:

Let's factor z2−10z+25

z2−10z+25

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(z+_)(z+_)

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Fill in the blanks in

(z+_)(z+_)

with -5 and -5 to get...

(z-5)(z-5)

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