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3 years ago

14

HELP ASAP I WILL GIVE BRAINLIEST FOR CORRECT ANSWER

2 answers:

wolverine [178]3 years ago

6 0

20

Hdhejejekkejenrnfjfj

yuradex [85]3 years ago

5 0

Answer:

20

Step-by-step explanation:

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8n - 7 = -5(-3n - 7)<br> Solve Multi-Step equation.<br> Please explain how you i can understand too!

Morgarella [4.7K]

Answer:

n=-6

Step-by-step explanation:

8n-7=-5(-3n-7)

Multiply -5 to -3n and -7

8n-7= 15n+35

Bring 8n to the other side by subracting it from 15n

-7=7n+35

Subtract 35 from both sides

-42=7n

Divide both sides by 7

n=-6

6 0

3 years ago

erica can run 1/6 of a kilometer in a minute her school is 3/4 of a kilometer away from her home at this speed how long would it

nlexa [21]

Answer: 4.5 minutes.

Step-by-step explanation:

1. You have the following information given in the problem above:

- Erica runs $\frac{1}{6}km$ in a minute.

- The school is $\frac{3}{4}km$ away from her home.

2. Therefore, to solve the exercise you must multiply $\frac{3}{4}km=0.75km$ by 1 minute and divide this by $\frac{1}{6}km=0.166km$ , as following:

$x=\frac{(0.75km)(1min)}{(0.166km)}\\x=4.5min$

3. Therefore, the result is 4.5 minutes.

7 0

3 years ago

Read 2 more answers

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

Here is an expression.

Zielflug [23.3K]

Answer:

sorry..

Step-by-step explanation:

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3 0

3 years ago

With the spinner shown, the theoretical probability of landing on 2 is 0.25. Which number of spins would most likely result in a

Mice21 [21]

just give the other kid brainiest

Step-by-step explanation:

5 0

3 years ago