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3 years ago

4% of what number is 40?

Mathematics

2 answers:

Gekata [30.6K]3 years ago

5 0

Answer: 1,000

Step-by-step explanation: Well percent means over 100 so we know that 4% means the same thing as 4/100. Of means times, what number means x, is means equals, and 40 means 40.

So we have the equation 4/100 · x = 40.

Notice that 4/100 can be reduced to 1/25.

So rewriting our problem we have x/25 = 40.

To get rid of the fraction, we multiply both sides of the equation by 25 and we have x = 1,000.

Now let's check our answer back in the

original problem to see if it makes sense.

We have 4% of 1,000 is 40.

Well we know that 100% of 1,000 would be 1,000 so 4% of 1,000 should be alot less than 1,000 so 40 seems to make sense.

I have also attached my own work in the image provided.

ladessa [460]3 years ago

4 0

Answer: 1000

Step-by-step explanation: you do 4% multiplied by x = 40

Multiplying both sides by 100 and dividing both sides by 4,

we have x = 40 × 100/4

40×100÷4

And you get 1000

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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

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

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

What is the answer for this please help -31=3w+9-8w simplify you answer as much as possible

ki77a [65]

$-31=3w+9-8w\\\\-5w+9=-31 \ \ /-9\\\\-5w=-40 \ \ /:(-5)\\\\\huge\boxed{w=8}$