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DiKsa [7]
4 years ago
6

60 POINTS!!! CAN SOMEBODY PLEASE HELP ME!! I DONT WANT TO FAIL THIS!

Mathematics
1 answer:
Kazeer [188]4 years ago
4 0

Answer:

why did it delete my answer

Step-by-step explanation:

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1. Solve the equation. (1 point)<br> 5x-1= 14
Rufina [12.5K]

Answer: x = 3

Step-by-step explanation:

5x - 1 + 14 - original equation

5x = 15 - add 1 to both sides

x = 3 - divide both sides by 5

4 0
3 years ago
24 POINTS!!!!!!!!!!!!!!!!!!
melomori [17]
9+9+8+4\pi

<span>4π+26 in.</span>
8 0
3 years ago
Read 2 more answers
You go to the mall to make some purchases. You buy 3 dresses which cost $59.99 each and a pair of shoes that cost $75. What is t
mina [271]
Combine costs
59.99(3) + 75 = 179.97 + 75
179.97 + 75 = $254.99 (total cost)
Now add tax: 8% = 0.08
254.99 * 0.08 = 20.3992
254.99 + 20.3992 = 275.3892

Solution: bill would be around $275
6 0
3 years ago
We would like to use the power series method to find the general solution to the differential equation d 2y dx2 − 4x dy dx + 12y
Feliz [49]

y=\displaystyle\sum_{n\ge0}a_nx^n

\dfrac{\mathrm dy}{\mathrm dx}=\displaystyle\sum_{n\ge1}na_nx^{n-1}\implies4x\dfrac{\mathrm dy}{\mathrm dx}=4\sum_{n\ge1}na_nx^n=4\sum_{n\ge0}na_nx^n

\dfrac{\mathrm d^2y}{\mathrm dx^2}=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}=\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n

Substituting into the ODE

\dfrac{\mathrm d^2y}{\mathrm dx^2}-4x\dfrac{\mathrm dy}{\mathrm dx}+12y=0

gives

\displaystyle\sum_{n\ge0}\bigg((n+2)(n+1)a_{n+2}-4na_n+12a_n\bigg)x^n=0

so that the coefficients of the series are given according to

\begin{cases}a_0=y(0)\\a_1=y'(0)\\a_{n+2}=\dfrac{4(n-3)a_n}{(n+2)(n+1)}&\text{for }n\ge0\end{cases}

We can shift the index in the recursive part of this definition to get

a_n=\dfrac{4(n-5)a_{n-2}}{n(n-1)}

for n\ge2. There's dependency between coefficients that are 2 indices apart, so we can consider 2 cases:

  • If n=2k, where k\ge0 is an integer, then

k=0\implies n=0\implies a_0=a_0

but since y(0)=0, we have a_0=0 and a_{2k}=0 for all k\ge0.

  • If n=2k+1, then

k=0\implies n=1\implies a_1=a_1

k=1\implies n=3\implies a_3=\dfrac{4(-2)a_1}{3\cdot2}=-\dfrac43a_1

k=2\implies n=5\implies a_5=0

and so a_{2k+1}=0 for all k\ge2. If y'(0)=1, we then have a_1=1 and a_3=-\dfrac43.

So the ODE has solution

y(x)=\displaystyle\sum_{k\ge0}(a_{2k}x^{2k}+a_{2k+1}x^{2k+1})\implies\boxed{y(x)=x-\dfrac43x^3}

8 0
3 years ago
-2x-y=0, where x≤0<br><br> Find the 6 trigonometric functions
Zina [86]

Answer:

very hard, why don't you search on internet

5 0
3 years ago
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