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fgiga [73]
3 years ago
11

Miranda bought a shirt for 36.99 dollar but it was for 33 percent off what is the original price?

Mathematics
1 answer:
boyakko [2]3 years ago
8 0
55.21 is the original price
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Simplify the difference quotient f(x)=x^2+9
Setler [38]

Answer:

sorry is its complicated

Step-by-step explanation:

Find the components of the definition.

f

(

x

+

h

)

=

x

2

+

2

h

x

+

h

2

−

9

f

(

x

)

=

x

2

−

9

Plug in the components.

f

(

x

+

h

)

−

f

(

x

)

h

=

x

2

+

2

h

x

+

h

2

−

9

−

(

x

2

−

9

)

h

8 0
3 years ago
Whats the gcf of 32 and 52
Molodets [167]
The greatest common factor of 32 and 52 is: 4

Factors of 32: 1, 2, 4, 8, 16, 32
Factors of 52: 1, 2, 4, 13, 26

The common factors between them are 1, 2, 4, but the greatest common factor between them is 4.


7 0
3 years ago
Read 2 more answers
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
Copy machine can make 250 copies in 4 minutes. Represent<br> rate of change in copies per minute.
pentagon [3]

Answer:

1000

Step-by-step explanation:

8 0
1 year ago
What is the surface area of the figure below
skad [1K]
To find the area, your just supposed to multiply
4 0
2 years ago
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