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2 years ago

what is the equation of the line that is perpendicular to the given line and has an x-intercept of 6?

Mathematics

1 answer:

yaroslaw [1]2 years ago

6 0

Answer:

y = 4/3x - 6

Step-by-step explanation:

The slope of the 1st line is -3/4

So for the perpendicular line it has to be the exact opposite.

That looks like this ---> 4/3x

And the y-intercept is -6 because if you make a straight line where the x intercept is it matches.

The equation would look like this

y = 4/3x - 6

Hope this helps! Have a blessed day!

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Quick!! using y= 5x + 12 find the slope of a parallel line

icang [17]

Answer:

y=5x-10

Step-by-step explanation:

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2 years ago

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three different are rung at intervals of 36 minutes 40 minutes and 48 minutes if they were rung together at 9 a.m. at what time

Lena [83]

To find the time in which all the bells ring together, we need to find the LCM of 36,40,48.

Prime factorization of 36=2×2×3×3

Prime factorization of 40=2×2×2×5

Prime factorization of 48=2×2×2×2×3

Hence, LCM of 36,40,48=2×2×2×2×3×3×5=720 seconds.

720seconds=

720

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Hence, all the bells will ring together after 12 mi

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2 years ago

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(x^2y+e^x)dx-x^2dy=0

klio [65]

It looks like the differential equation is

$\left(x^2y + e^x\right) \,\mathrm dx - x^2\,\mathrm dy = 0$

Check for exactness:

$\dfrac{\partial\left(x^2y+e^x\right)}{\partial y} = x^2 \\\\ \dfrac{\partial\left(-x^2\right)}{\partial x} = -2x$

As is, the DE is not exact, so let's try to find an integrating factor µ(x, y) such that

$\mu\left(x^2y + e^x\right) \,\mathrm dx - \mu x^2\,\mathrm dy = 0$

*is* exact. If this modified DE is exact, then

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \dfrac{\partial\left(-\mu x^2\right)}{\partial x}$

We have

$\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu$

Notice that if we let µ(x, y) = µ(x) be independent of y, then ∂µ/∂y = 0 and we can solve for µ :

$x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}$

The modified DE,

$\left(e^{-x}y + \dfrac1{x^2}\right) \,\mathrm dx - e^{-x}\,\mathrm dy = 0$

is now exact:

$\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}$

So we look for a solution of the form F(x, y) = C. This solution is such that

$\dfrac{\partial F}{\partial x} = e^{-x}y + \dfrac1{x^2} \\\\ \dfrac{\partial F}{\partial y} = e^{-x}$

Integrate both sides of the first condition with respect to x :

$F(x,y) = -e^{-x}y - \dfrac1x + g(y)$

Differentiate both sides of this with respect to y :

$\dfrac{\partial F}{\partial y} = -e^{-x}+\dfrac{\mathrm dg}{\mathrm dy} = e^{-x} \\\\ \implies \dfrac{\mathrm dg}{\mathrm dy} = 0 \implies g(y) = C$

Then the general solution to the DE is

$F(x,y) = \boxed{-e^{-x}y-\dfrac1x = C}$

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3 years ago

Please help! If GDEF is a square, GE=4x+1 and DF=8x-51, find the value of x.

zepelin [54]

Answer:

x = 13

Step-by-step explanation:

1. 4x + 1 = 8x - 51

2. 4x = 8x - 52

3. -4x = -52

4. x = 13

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3 years ago

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App 19.study

lions [1.4K]

Answer:

$1 \to 22 \to 0.176$

$2 \to 13 \to 0.104$

$3 \to 18 \to 0.144$

$4 \to 29 \to 0.232$

$5 \to 37 \to 0.296$

$6 \to 6 \to 0.048$

Step-by-step explanation:

Given

$n = 125$

See attachment for proper table

Required

Complete the table

Experimental probability is calculated as:

$Pr = \frac{Frequency}{n}$

We use the above formula when the frequency is known.

For result of roll 2, 4 and 6

The frequencies are 13, 29 and 6, respectively

So, we have:

$Pr(2) = \frac{13}{125} = 0.104$

$Pr(4) = \frac{29}{125} = 0.232$

$Pr(6) = \frac{6}{125} = 0.048$

When the frequency is to be calculated, we use:

$Pr = \frac{Frequency}{n}$

$Frequency = n * Pr$

For result of roll 3 and 5

The probabilities are 0.144 and 0.296, respectively

So, we have:

$Frequency(3) = 125 * 0.144 = 18$

$Frequency(5) = 125 * 0.296 = 37$

For roll of 1 where the frequency and the probability are not known, we use:

$Total \ Frequency = 125$

So:

Frequency(1) added to others must equal 125

This gives:

$Frequency(1) + 13 + 18 + 29 + 37 + 6 = 125$

$Frequency(1) + 103 = 125$

Collect like terms

$Frequency(1) =- 103 + 125$

$Frequency(1) =22$

The probability is then calculated as:

$Pr(1) = \frac{22}{125}$

$Pr(1) = 0.176$

So, the complete table is:

$1 \to 22 \to 0.176$

$2 \to 13 \to 0.104$

$3 \to 18 \to 0.144$

$4 \to 29 \to 0.232$

$5 \to 37 \to 0.296$

$6 \to 6 \to 0.048$

5 0

2 years ago

what is the equation of the line that is perpendicular to the given line and has an x-intercept of 6?​

what is the equation of the line that is perpendicular to the given line and has an x-intercept of 6?