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Diano4ka-milaya [45]
3 years ago
8

BD bisects ABC. Find the value of x. a.x=3/17 b.x=3 c.x=9 d.x=-3

Mathematics
1 answer:
wolverine [178]3 years ago
8 0
The answer is option ‘b’

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Perform the indicated operation.<br> (c 2 + 2c - 35) ÷ (c - 5)
deff fn [24]
(c²+2c-35): (c-5)
Ruffini´s rule:

             1              2              -35

5                           5                35  
-------------------------------------------------
             1              7                  0


(c²+2c-35): (c-5)=(c+7)

4 0
3 years ago
Read 2 more answers
14+5/7x-20=4(1/7x-1)
katrin2010 [14]
14-20= - 6
5/7x-6=4/7x-4
5/7x-4/7x= 1/7x
1/7x-6=-4
-4+6= 2
1/7x divide by 2= so x= 14
7 0
3 years ago
Science Final Exam
Olin [163]

Answer:

  24

Step-by-step explanation:

All of the students in the last three grade groups scored 70% or better:

  11 + 8 + 5 = 24

24 students scored 70% or better.

8 0
3 years ago
1 point<br> What is the value of x?<br> 25°<br> Typo your answer<br> tº<br> 40°
spin [16.1K]

....?.,,ekskkdksksks

8 0
2 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
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