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3 years ago

A line has slope –5(over)3 . Through which two points could this line pass? (1 point)

Mathematics

2 answers:

hoa [83]3 years ago

5 0

Answer: Rate of Change and Slope Quick Check

C. -1/3

D. (11,13) (8,18)

B. -2

D. Undefined

D. 55/1

Step-by-step explanation:

motikmotik3 years ago

3 0

The points of the line that has a slope of -5 over 3 is (11,13) (8, 18). This can be computed using the formula of finding the slope of a line which m = y2-y2 over x2-x1. This is computed as follows:

m = 18-13 over 8 - 11
m= 5 over -3 or simplified to m= -5 over 3

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$x^2y_1v''+(2x^2{y_1}'-3xy_1)v'+(x^2{y_1}''-3x{y_1}'+5y_1)v=0$

$x^4\cos(\ln x)v''+x^3(\cos(\ln x)-2\sin(\ln x))v'=0$

which leaves us with an ODE linear in $w(x)=v'(x)$ :

$x^4\cos(\ln x)w'+x^3(\cos(\ln x)-2\sin(\ln x))w=0$

This ODE is separable; divide both sides by the coefficient of $w'(x)$ and separate the variables to get

$w'+\dfrac{\cos(\ln x)-2\sin(\ln x)}{x\cos(\ln x)}w=0$

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$\dfrac{\mathrm dw}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}\,\mathrm dx$

Integrate both sides; on the right, substitute $u=\ln x$ so that $\mathrm du=\dfrac{\mathrm dx}x$ .

$\ln|w|=\displaystyle\int\frac{2\sin u-\cos u}{\cos u}\,\mathrm du=\int(2\tan u-1)\,\mathrm du$

Now solve for $w(u)$ ,

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$w=e^{-2\ln(\cos u)-u+C}$

$w=Ce^{-u}\sec^2u$

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$w=Ce^{-\ln x}\sec^2(-\ln x)$

$w=C\dfrac{\sec^2(\ln x)}x$

Solve for $v(x)$ by integrating both sides.

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Substitute $u=\ln x$ again and solve for $v(u)$ :

$v=\displaystyle C_1\int\sec^2u\,\mathrm du$

$v=C_1\tan u+C_2$

then for $v(x)$ ,

$v=C_1\tan(\ln x)+C_2$

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$y_2=x^2\cos(\ln x)(C_1\tan(\ln x)+C_2)$

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