All categories

3 years ago

The sum of the ages of Nicole and Rachel is 30. Currently Nicole is four years older than Rachel. How old is Rachel?

Mathematics

2 answers:

PSYCHO15rus [73]3 years ago

7 0

Since Nicole is 17, and Rachel is four years younger, her age is 13. This answer makes sense because 17+13=30.

Lunna [17]3 years ago

3 0

Rachel is 13. since Nicole is 17.
13+17=30

You might be interested in

I literally don’t get this I need the answer

Ksju [112]

Answer:

62.83 $units^{3}$

Step-by-step explanation:

The equation is V(Volume) equals π (Pi/3.14) multiplied by r(radius) squared multiplied by h(height). The way it would be written out is V=π× $r^{2}$ ×h. So when you plug in the radius (2) and the height (5) you get π× $2^{2}$ ×5 which equals 62.83185 or 62.83. I hope this helps you understand and if you need more help let me know.

6 0

3 years ago

What is the value of S in the equation 1/3s+14=26

Law Incorporation [45]

First, isolate 1/3s:
1/3s=12
Second, multiply both sides by 3 to get a singular value of s (just s):
s=36

4 0

3 years ago

1. Choose the best answer that represents the property used to rewrite the expression.

Likurg_2 [28]

in log minus means divide

log 6 - log 3 = log (6÷3)

log 6 - log 3 = log 2

6 0

2 years ago

If the length of a rectangle is 5a^2 and the width is 7ab^2 what is the area of the rectangle?

DanielleElmas [232]

Answer:

I think u forgot to put a photo because I really don't understand

Step-by-step explanation:

not sure

8 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