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vaieri [72.5K]
3 years ago
7

What is 2 3/15 roumded to the nearest half

Mathematics
1 answer:
Black_prince [1.1K]3 years ago
5 0

Answer:

It is 7 whole

Step-by-step explanation:

3/15 is 5 without anything left plus the 3 is 7

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Number Lines / Algebra. Can anyone please tell me to answers to all of these number lines below. Thank you!!
iren [92.7K]
I will do the first-two lines only.

-4 -3 -2 -1 0 1 2 3 4 5 6

Next line below.

-5 -4 -3 -2 -1 0 1 2 3 4 5

Do the rest.
5 0
3 years ago
What is the risk of going into a group of more than 20 or more people at this time from coronavirus?
olga55 [171]

Answer:

Even if a person doesn't show symptoms, they can still have it.

Step-by-step explanation:

That's the problem.

Even if you think everyone isn't sick, they very well could be, and because there is no vaccine and it spreads so quickly, there's a chance that all the people in that group could get it.

5 0
3 years ago
Read 2 more answers
__Na2O + __ H2O → __NaOH
Andrews [41]

Answer:

Na20+H20 -> 2NaOH

Step-by-step explanation:

8 0
3 years ago
An oil company fills 1 over 12 of a tank in 1 over 3 hour. At this rate, which expression can be used to determine how long will
Leto [7]
Don't let fractions fool you.

\frac{1}{12} of a tank in \frac{1}{3} an hour is equal to:
1/12 of a tank in 20 minutes.

Thus we know in an hour, an oil company can fill 3/12 of a tank (60 minutes in an hour).

3 x 4 = 12, and 12/12 = 1(a whole tank), so we multiply 1 (hour) by 4.

Our answer is that it takes 4 hours to fill a tank.

6 0
3 years ago
Find two power series solutions of the given differential equation about the ordinary point x = 0. compare the series solutions
monitta
I don't know what method is referred to in "section 4.3", but I'll suppose it's reduction of order and use that to find the exact solution. Take z=y', so that z'=y'' and we're left with the ODE linear in z:

y''-y'=0\implies z'-z=0\implies z=C_1e^x\implies y=C_1e^x+C_2

Now suppose y has a power series expansion

y=\displaystyle\sum_{n\ge0}a_nx^n
\implies y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}
\implies y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}

Then the ODE can be written as

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge1}na_nx^{n-1}=0

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}-\sum_{n\ge2}(n-1)a_{n-1}x^{n-2}=0

\displaystyle\sum_{n\ge2}\bigg[n(n-1)a_n-(n-1)a_{n-1}\bigg]x^{n-2}=0

All the coefficients of the series vanish, and setting x=0 in the power series forms for y and y' tell us that y(0)=a_0 and y'(0)=a_1, so we get the recurrence

\begin{cases}a_0=a_0\\\\a_1=a_1\\\\a_n=\dfrac{a_{n-1}}n&\text{for }n\ge2\end{cases}

We can solve explicitly for a_n quite easily:

a_n=\dfrac{a_{n-1}}n\implies a_{n-1}=\dfrac{a_{n-2}}{n-1}\implies a_n=\dfrac{a_{n-2}}{n(n-1)}

and so on. Continuing in this way we end up with

a_n=\dfrac{a_1}{n!}

so that the solution to the ODE is

y(x)=\displaystyle\sum_{n\ge0}\dfrac{a_1}{n!}x^n=a_1+a_1x+\dfrac{a_1}2x^2+\cdots=a_1e^x

We also require the solution to satisfy y(0)=a_0, which we can do easily by adding and subtracting a constant as needed:

y(x)=a_0-a_1+a_1+\displaystyle\sum_{n\ge1}\dfrac{a_1}{n!}x^n=\underbrace{a_0-a_1}_{C_2}+\underbrace{a_1}_{C_1}\displaystyle\sum_{n\ge0}\frac{x^n}{n!}
4 0
3 years ago
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