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amid [387]
3 years ago
14

A line is drawn through (–7, 11) and (8, –9). The equation y – 11 = y minus 11 equals StartFraction negative 4 Over 3 EndFractio

n left-parenthesis x plus 7 right-parenthesis.(x + 7) is written to represent the line. Which equations also represent the line? Check all that apply.
y = y equals StartFraction negative 4 Over 3 EndFraction left-parenthesis x plus StartFraction 5 Over 3 EndFraction.x +
3y = –4x + 40
4x + y = 21
4x + 3y = 5
–4x + 3y = 17
Mathematics
2 answers:
yaroslaw [1]3 years ago
6 0

⇒Given equation of line Passing through  (–7, 11) and (8, –9) is given by

       y-11= \frac{-4}{3}(x+7)

⇒Equation of line Passing through  (–7, 11) and (8, –9) is given by

   \rightarrow \frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\\rightarrow \frac{y-11}{x-(-7)}=\frac{-9-11}{8-(-7)}\\\\\rightarrow \frac{y-11}{x+7}=\frac{-20}{15}\\\\\rightarrow \frac{y-11}{x+7}=\frac{-4}{3}\\\\\rightarrow 3y-33=-4x-28\\\\\rightarrow 4x+3y=33-28\\\\ \rightarrow 4x+3y=5

Option C

          4x+3y=5

Pachacha [2.7K]3 years ago
5 0

Answer:

A and C

Step-by-step explanation:

Shanice
3 years ago
It’s Aand D
Naetoosmart
2 years ago
no its not they were right
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Hello

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4 years ago
The base of a triangle is 3.4cm.the area is 20.74cm.find the height to 1 dp​
inessss [21]

Answer:

12.2 cm

Step-by-step explanation:

The area (A) of a triangle is calculated as

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Here A = 20.74 and b = 3.4 , thus

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3 years ago
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Can you help me understand this question​
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Answer:

Which one?

Step-by-step explanation:

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For the differential equation 3x^2y''+2xy'+x^2y=0 show that the point x = 0 is a regular singular point (either by using the lim
Svetlanka [38]
Given an ODE of the form

y''(x)+p(x)y'(x)+q(x)y(x)=f(x)

a regular singular point x=c is one such that p(x) or q(x) diverge as x\to c, but the limits of (x-c)p(x) and (x-c)^2q(x) as x\to c exist.

We have for x\neq0,

3x^2y''+2xy'+x^2y=0\implies y''+\dfrac2{3x}y'+\dfrac13y=0

and as x\to0, we have x\cdot\dfrac2{3x}\to\dfrac23 and x^2\cdot\dfrac13\to0, so indeed x=0 is a regular singular point.

We then look for a series solution about the regular singular point x=0 of the form

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

Substituting into the ODE gives

\displaystyle3x^2\sum_{n\ge0}a_n(n+k)(n+k-1)x^{n+k-2}+2x\sum_{n\ge0}a_n(n+k)x^{n+k-1}+x^2\sum_{n\ge0}a_nx^{n+k}=0

\displaystyle3\sum_{n\ge2}a_n(n+k)(n+k-1)x^{n+k}+3a_1k(k+1)x^{k+1}+3a_0k(k-1)x^k
\displaystyle+2\sum_{n\ge2}a_n(n+k)x^{n+k}+2a_1(k+1)x^{k+1}+2a_0kx^k
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From this we find the indicial equation to be

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Taking k=\dfrac13, and in the x^{k+1} term above we find a_1=0. So we have

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Since a_1=0, all coefficients with an odd index will also vanish.

So the first three terms of the series expansion of this solution are

\displaystyle\sum_{n\ge0}a_nx^{n+1/3}=a_0x^{1/3}+a_2x^{7/3}+a_4x^{13/3}

with a_0=1, a_2=-\dfrac1{14}, and a_4=\dfrac1{728}.
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Answer:

Step-by-step explanation:

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