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3 years ago

Write an equation that represents the line use exact numbers

Mathematics

1 answer:

RSB [31]3 years ago

3 0

Answer:

y = (-6/7)x - 1

Step-by-step explanation:

There's a dot where this line crosses the y-axis; it's represented by the point (0, -1). Consequently, the y-intercept, b, of this line is -1.

As x increases from -6 to 0, y decreases from 7 to -1. The change in x is 7 units and the change in y is -6 units. Thus, the slope of this line is

m = rise / run = -6 / 7.

Again, the y-intercept is b = -1.

Thus, inserting the knowns into the slope-intercept form of the equation of a straight line results in:

y = (-6/7)x - 1

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Aleks [24]

Answer:

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Step-by-step explanation:

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3 years ago

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Nikolay [14]

Replace x with π/2 - x to get the equivalent integral

$\displaystyle \int_{-\frac\pi2}^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

but the integrand is even, so this is really just

$\displaystyle 2 \int_0^{\frac\pi2} \cos(\cot(x) - \tan(x)) \, dx$

Substitute x = 1/2 arccot(u/2), which transforms the integral to

$\displaystyle 2 \int_{-\infty}^\infty \frac{\cos(u)}{u^2+4} \, du$

There are lots of ways to compute this. What I did was to consider the complex contour integral

$\displaystyle \int_\gamma \frac{e^{iz}}{z^2+4} \, dz$

where γ is a semicircle in the complex plane with its diameter joining (-R, 0) and (R, 0) on the real axis. A bound for the integral over the arc of the circle is estimated to be

$\displaystyle \left|\int_{z=Re^{i0}}^{z=Re^{i\pi}} f(z) \, dz\right| \le \frac{\pi R}{|R^2-4|}$

which vanishes as R goes to ∞. Then by the residue theorem, we have in the limit

$\displaystyle \int_{-\infty}^\infty \frac{\cos(x)}{x^2+4} \, dx = 2\pi i {} \mathrm{Res}\left(\frac{e^{iz}}{z^2+4},z=2i\right) = \frac\pi{2e^2}$

and it follows that

$\displaystyle \int_0^\pi \cos(\cot(x)-\tan(x)) \, dx = \boxed{\frac\pi{e^2}}$

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2 years ago

Which of the following illustrates a phase shift?

Natasha_Volkova [10]

Answer:

A. y = -2 - cos(x-π)

Explanation:

The general form of the trig equation is:

y = A sin (Bx + C) + D

where:

A is the amplitude

$\frac{2\pi}{B}$ is the period

$\frac{-C}{B}$ is the phase shift

D is the vertical shift

Now, let's check the choices:

A. y = -2 - cos(x-π)

$\frac{-C}{B} = \frac{\pi}{1} = \pi$

Therefore, the function has a phase shift of π

B. y = 3 cos(4x)

$\frac{-C}{B} = \frac{0}{4} = 0$

Therefore, the function has no phase shift

C. y = tan(2x)

$\frac{-C}{B} = \frac{0}{2} = 0$

Therefore, the function has no phase shift

D. y = 1 + sin(x)

$\frac{-C}{B} = \frac{0}{1} = 0$

Therefore, the function has no phase shift

Based on the above, the correct answer is A

Hope this helps :)

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3 years ago

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Jane solved an equation and got a solution of X=3. The teacher then asked what the value of 6X would be. Which of the answer is

weeeeeb [17]

If the solution of x is 3, then the solution of 6x would be 18, because 3*6=18. :)

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3 years ago

A fraction equal to one for the relationship between these units m/c=1

BigorU [14]

$\dfrac{m}{c}=1\ \text{therefore}\ m=c\ and\ m,c\neq0$

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3 years ago