Can you please help answer this

Give an example of two irrational numbers whose a) difference is an

astra-53 [7]

Answer:

2V3 and V3

Step-by-step explanation:

a) 2V3-V3=V3 (irational)

b) 2V3*V3=2*3=6 (rational)

c) 2V3/V3=2 (rational)

4 0

3 years ago

There are 12300 films on Netflix. 7% of the films are action. How many films are action?

Kaylis [27]

Answer:

861

Step-by-step explanation:

7 percent *12300 =

(7:100)*12300 =

(7*12300):100 =

86100:100 = 861

7 0

3 years ago

13.10. Suppose that a sequence (ao, a1, a2, ) of real numbers satisfies the recurrence relation an -5an-1+6an-20 for all n> 2

Gwar [14]

a. This recurrence is of order 2.

b. We're looking for a function $A(x)$ such that

$A(x)=\displaystyle\sum_{n=0}^\infty a_nx^n$

Take the recurrence,

$\begin{cases}a_0=a_0\\a_1=a_1\\a_n-5a_{n-1}+6a_{n-2}=0&\text{for }n\ge2\end{cases}$

Multiply both sides by $x^{n-2}$ and sum over all integers $n\ge2$ :

$\displaystyle\sum_{n=2}^\infty a_nx^{n-2}-5\sum_{n=2}^\infty a_{n-1}x^{n-2}+6\sum_{n=2}^\infty a_{n-2}x^{n-2}=0$

Pull out powers of $x$ so that each summand takes the form $a_kx^k$ :

$\displaystyle\frac1{x^2}\sum_{n=2}^\infty a_nx^n-\frac5x\sum_{n=2}^\infty a_{n-1}x^{n-1}+6\sum_{n=2}^\infty a_{n-2}x^{n-2}=0$

Now shift the indices and add/subtract terms as needed to get everything in terms of $A(x)$ :

$\displaystyle\frac1{x^2}\left(\sum_{n=0}^\infty a_nx^n-a_0-a_1x\right)-\frac5x\left(\sum_{n=0}^\infty a_nx^n-a_0\right)+6\sum_{n=0}^\infty a_nx^n=0$

$\displaystyle\frac{A(x)-a_0-a_1x}{x^2}-\frac{5(A(x)-a_0)}x+6A(x)=0$

Solve for $A(x)$ :

$A(x)=\dfrac{a_0+(a_1-5a_0)x}{1-5x+6x^2}\implies\boxed{A(x)=\dfrac{a_0+(a_1-5a_0)x}{(1-3x)(1-2x)}}$

c. Splitting $A(x)$ into partial fractions gives

$A(x)=\dfrac{2a_0-a_1}{1-3x}+\dfrac{3a_0-a_1}{1-2x}$

Recall that for $|x|$ , we have

$\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n$

so that for $|3x|$ and $|2x|$ , or simply $|x|$ , we have

$A(x)=\displaystyle\sum_{n=0}^\infty\bigg((2a_0-a_1)3^n+(3a_0-a_1)2^n\bigg)x^n$

which means the solution to the recurrence is

$\boxed{a_n=(2a_0-a_1)3^n+(3a_0-a_1)2^n}$

d. I guess you mean $a_0=2$ and $a_1=5$ , in which case

$\boxed{\begin{cases}a_0=2\\a_1=5\\a_2=13\\a_3=35\\a_4=97\\a_5=275\end{cases}}$

e. We already know the general solution in terms of $a_0$ and $a_1$ , so just plug them in:

$\boxed{a_n=2^n+3^n}$

8 0

3 years ago

HELPPPP PLEASEE

klio [65]

Answer:

B

Step-by-step explanation:

Using the graph, we pick a convenient point: (20, 180)

This point means that for 20 hours of work, Vincent is paid 180 dollars. In fraction form, we can write 20hrs/180 dollars. If we simplify the fraction by dividing both the numerator and denominator by 20 we get 1/9 AKA 1hr/9 dollars. So in 1 hour, Vinnie is paid $9.00

Have a good rest of your day

6 0

3 years ago

Read 2 more answers

HELP ME PLSSSSSS<br><br> if f(x) = 2x-3/5 , which of the following is the inverse of f(x)?

neonofarm [45]

The answer is (5x+3)/2 . So it’s A .

8 0

3 years ago

Can you please help answer this ​

Can you please help answer this