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Andreyy89
1 year ago
12

What is 8.7352 rounded to the nearest tenth?

Mathematics
1 answer:
MrMuchimi1 year ago
7 0
8.7352 rounded to the nearest tenth is: 8.7, because the tens place is 0.7 in this situation and looking at the next number, 3 is less than 5 so when rounding you keep the 7 the same.
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Paul is going to have the exterior of his home painted. Painting USA charges $75.00 to come out and evaluate the house plus $25.
murzikaleks [220]
CP - the cost of painting the house for Painting USA company [$]
cP = 75 + 25 * 25 = 75 + 625 = 700 [$]
cU - the cost of painting the house for Upscale Home Painting company [$]
cU = 0 + 40 * 25 = 1000 [$]

cP < cU
The most cost effective is Painting USA company.

n - number of houres required to paint the house if the same cost for both companies
75 + 25 * n = 40 * n
40 n - 25n = 75
15n = 75  /:15
n = 5
The cost of paining the house is the same for both companies if the number of houres required to paint the house is 5.


6 0
2 years ago
NL and MK are diameters of circle J. Complete the statements about circle J. MP is a major arc/minor arc and measures __ degrees
joja [24]

Answer:

1. Minor arc

2. 152

3. major arc

4. 298

Step-by-step explanation:

3 0
2 years ago
To the nearest foot, what is the width of the arch 8 ft above the water?
murzikaleks [220]

Answer:

15 is the answer on edge

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
Power series of y''+x^2y'-xy=0
Ray Of Light [21]
Assuming we're looking for a power series solution centered around x=0, take

y=\displaystyle\sum_{n\ge0}a_nx^n
y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}
y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}

Substituting into the ODE yields

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}+\sum_{n\ge1}na_nx^{n+1}-\sum_{n\ge0}a_nx^{n+1}=0

The first series starts with a constant term; the second series starts at x^2; the last starts at x^1. So, extract the first two terms from the first series, and the first term from the last series so that each new series starts with a x^2 term. We have

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}=2a_2+6a_3x+\sum_{n\ge4}n(n-1)a_nx^{n-2}

\displaystyle\sum_{n\ge0}a_nx^{n+1}=a_0x+\sum_{n\ge1}a_nx^{n+1}

Re-index the first sum to have it start at n=1 (to match the the other two sums):

\displaystyle\sum_{n\ge4}n(n-1)a_nx^{n-2}=\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}

So now the ODE is

\displaystyle\left(2a_2+6a_3x+\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}\right)+\sum_{n\ge1}na_nx^{n+1}-\left(a_0x+\sum_{n\ge1}a_nx^{n+1}\right)=0

Consolidate into one series starting n=1:

\displaystyle2a_2+(6a_3-a_0)x+\sum_{n\ge1}\bigg[(n+3)(n+2)a_{n+3}+(n-1)a_n\bigg]x^{n+1}=0

Suppose we're given initial conditions y(0)=a_0 and y'(0)=a_1 (which follow from setting x=0 in the power series representations for y and y', respectively). From the above equation it follows that

\begin{cases}2a_2=0\\6a_3-a_0=0\\(n+3)(n+2)a_{n+3}+(n-1)a_n=0&\text{for }n\ge2\end{cases}

Let's first consider what happens when n=3k-2, i.e. n\in\{1,4,7,10,\ldots\}. The recurrence relation tells us that

a_4=-\dfrac{1-1}{(1+3)(1+2)}a_1=0\implies a_7=0\implies a_{10}=0

and so on, so that a_{3k-2}=0 except for when k=1.

Now let's consider n=3k-1, or n\in\{2,5,8,11,\ldots\}. We know that a_2=0, and from the recurrence it follows that a_{3k-1}=0 for all k.

Finally, take n=3k, or n\in\{0,3,6,9,\ldots\}. We have a solution for a_3 in terms of a_0, so the next few terms (k=2,3,4) according to the recurrence would be

a_6=-\dfrac2{6\cdot5}a_3=-\dfrac2{6\cdot5\cdot3\cdot2}a_0=-\dfrac{a_0}{6\cdot3\cdot5}
a_9=-\dfrac5{9\cdot8}a_6=\dfrac{a_0}{9\cdot6\cdot3\cdot8}
a_{12}=-\dfrac8{12\cdot11}a_9=-\dfrac{a_0}{12\cdot9\cdot6\cdot3\cdot11}

and so on. The reordering of the product in the denominator is intentionally done to make the pattern clearer. We can surmise the general pattern for n=3k as

a_{3k}=\dfrac{(-1)^{k+1}a_0}{(3k\cdot(3k-3)\cdot(3k-2)\cdot\cdots\cdot6\cdot3\cdot(3k-1)}
a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^k(k\cdot(k-1)\cdot\cdots\cdot2\cdot1)\cdot(3k-1)}
a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^kk!(3k-1)}

So the series solution to the ODE is given by

y=\displaystyle\sum_{n\ge0}a_nx^n
y=a_1x+\displaystyle\sum_{k\ge0}\frac{(-1)^{k+1}a_0}{3^kk!(3k-1)}

Attached is a plot of a numerical solution (blue) to the ODE with initial conditions sampled at a_0=y(0)=1 and a_1=y'(0)=2 overlaid with the series solution (orange) with n=3 and n=6. (Note the rapid convergence.)

7 0
2 years ago
Find the equation of the line below if necessary use the slash (/) to indicate a division bar (2,6)(1,3)
777dan777 [17]
Given that the line passes through points:
(2,6), (1,3)
the equation will be given by:
m(x-x1)=y-y1
where
slope, m=(y-y1)/(x-x1)
from the points given:
m=(6-3)/(2-1)=3/1=3
thus the equation will be given by:
using point (1,3)
3(x-1)=y-3
3x-3=y-3
hence
y=3x-3+3
y=3x
Answer: y=3x
7 0
2 years ago
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