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svp [43]
3 years ago
11

Graph ​y=−47x+1​. Use the line tool and select two points on the line.

Mathematics
1 answer:
ratelena [41]3 years ago
8 0

Answer:

Ok so I don't have a graphing thing on my computer but on the y axis you are going to put a dot on the one then go up 47 and to the right 1 and put a dot sorry that is as much as i can help you

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What is the surface area for the figure? PLEASE HELP
olga_2 [115]

Answer:

516 cm squared

Step-by-step explanation:

To find the surface area of a rectangular prism such as this, is basically just like finding the OUTSIDE surfaces. We are looking for the "perimeter" of a 3D shape.

How do we do it?

  1. Calculate each face of the prism by multiplying Length x Width.
  2. Make a shortcut by combining like-figures (front of the cube and back of the cube are the same surface areas, etc.)
  3. Make your formula

  • 2(4 · 15) + 2(10 · 15) + 2(12 · 4)

6 0
3 years ago
Using power series, solve the LDE: (2x^2 + 1) y" + 2xy' - 4x² y = 0 --- - -- -
sattari [20]

We're looking for a solution of the form

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

with derivatives

y'=\displaystyle\sum_{n\ge0}(n+1)a_{n+1}x^n

y''=\displaystyle\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n

Substituting these into the ODE gives

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

Shifting indices to get each term in the summand to start at the same power of x and pulling the first few terms of the resulting shifted series as needed gives

2a_2+(2a_1+6a_3)x+\displaystyle\sum_{n\ge2}\bigg((n+2)(n+1)a_{n+2}+2n^2a_n-4a_{n-2}\bigg)x^n=0

Then the coefficients in the series solution are given according to the recurrence

\begin{cases}a_0=y(0)\\\\a_1=y'(0)\\\\a_2=0\\\\2a_1+6a_3=0\implies a_3=-\dfrac{a_1}3\\\\a_n=\dfrac{-2(n-2)^2a_{n-2}+4a_{n-4}}{n(n-1)}&\text{for }n\ge4\end{cases}

Given the complexity of this recursive definition, it's unlikely that you'll be able to find an exact solution to this recurrence. (You're welcome to try. I've learned this the hard way on scratch paper.) So instead of trying to do that, you can compute the first few coefficients to find an approximate solution. I got, assuming initial values of y(0)=y'(0)=1, a degree-8 approximation of

y(x)\approx1+x-\dfrac{x^3}3+\dfrac{x^4}3+\dfrac{x^5}2-\dfrac{16x^6}{45}-\dfrac{79x^7}{125}+\dfrac{101x^8}{210}

Attached are plots of the exact (blue) and series (orange) solutions with increasing degree (3, 4, 5, and 65) and the aforementioned initial values to demonstrate that the series solution converges to the exact one (over whichever interval the series converges, that is).

5 0
3 years ago
How much water do 4 ice cubes have?
lisov135 [29]
Found this on the internet hope it helps

I found that the average ice cube has two tablespoon or one ounce in one ice cube so 4 ice cubes will have 4 ounces
3 0
3 years ago
The first term of the sequence is 13 the term to rule is take away -5 work out the 4th term
Zigmanuir [339]

1st =13

2nd = 13- -5  = 13+5 =18

3rd = 18 - -5 = 18+5 =23

4th = 23 - -5 = 23+5 = 28

8 0
4 years ago
How many of each angles are in a pentagon
Illusion [34]

There are 108 degrees in each interior angle of a regular pentagon .


I hope that's help !

8 0
4 years ago
Read 2 more answers
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