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2 years ago

9

HELP ASAP!!!!!!!!! Please answer both

1 answer:

maks197457 [2]2 years ago

3 0

Step-by-step explanation:

questions 8, obviously the mid part of the parabola is 3, that is the axis of symmetry.

question 9, only one solution because you only have one parabola.

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What is the equation, in point slope form, that goes through (-8, 1) and has a slope of 5/6

yawa3891 [41]

First, solve for slope.
Formula for slope:
Plug in values:
Slope:
Second, we must plug into point-slope form.
Formula for point-slope:
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5 0

2 years ago

What is the equation of the line that is perpendicular to the given line and passes through the point (3, 0)?

olga2289 [7]

Step $1$

<u>Find the slope of the given line</u>

Let

$A(-3,2)\ B(2,-1)$

slope mAB is equal to

$mAB=\frac{(y2-y1)}{(x2-x1)} \\ \\ mAB=\frac{(-1-2)}{(2+3)} \\ \\ mAB=-\frac{3}{5}$

Step $2$

<u>Find the slope of the line that is perpendicular to the given line</u>

Let

CD ------> the line that is perpendicular to the given line

we know that

If two lines are perpendicular, then the product of their slopes is equal to $-1$

so

$mAB*mCD=-1\\ mAB=-\frac{3}{5} \\ mCD=-\frac{1}{mAB} \\ mCD=\frac{5}{3}$

Step $3$

<u>Find the equation of the line with mCD and the point (3,0)</u>

we know that

the equation of the line in the form point-slope is equal to

$y-y1=m(x-x1)\\\\ y-0=\frac{5}{3} *(x-3)\\\\ y=\frac{5}{3} x-5$

Multiply by $3$ both sides

$3y=5x-15$

$5x-3y=15$

therefore

the answer is

the equation of the line that is perpendicular to the given line is the equation $5x-3y=15$

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3 years ago

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Step-by-step explanation:

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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The general solution of 2 y ln(x)y' = (y^2 + 4)/x is

Sav [38]

Replace $y'$ with $\dfrac{\mathrm dy}{\mathrm dx}$ to see that this ODE is separable:

$2y\ln x\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{y^2+4}x\implies\dfrac{2y}{y^2+4}\,\mathrm dy=\dfrac{\mathrm dx}{x\ln x}$

Integrate both sides; on the left, set $u=y^2+4$ so that $\mathrm du=2y\,\mathrm dy$ ; on the right, set $v=\ln x$ so that $\mathrm dv=\dfrac{\mathrm dx}x$ . Then

$\displaystyle\int\frac{2y}{y^2+4}\,\mathrm dy=\int\dfrac{\mathrm dx}{x\ln x}\iff\int\frac{\mathrm du}u=\int\dfrac{\mathrm dv}v$

$\implies\ln|u|=\ln|v|+C$

$\implies\ln(y^2+4)=\ln|\ln x|+C$

$\implies y^2+4=e^{\ln|\ln x|+C}$

$\implies y^2=C|\ln x|-4$

$\implies y=\pm\sqrt{C|\ln x|-4}$

4 0

2 years ago