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umka2103 [35]
3 years ago
14

Consider the line 7x+9y=5

Mathematics
1 answer:
alisha [4.7K]3 years ago
4 0

Answer:

The slope of a line parallel to this line will be: -7/9

The slope of the perpendicular line will be:

\frac{-1}{\frac{-7}{9}}=\frac{9}{7}

Step-by-step explanation:

We know the slope-intercept form

y=mx+b

Here,

  • m is the slope
  • b is the y-intercept

Given the equation

7x+9y=5

simplifying to write in the lope-intercept form

y=-\frac{7}{9}x+\frac{5}{9}

Thus, the slope of the line is: -7/9

The slope of a line parallel to the line:

We have already determined that the slope of the line is: -7/9

  • We know that the parallel lines have the same slope.

Thus, the slope of a line parallel to this line will be: -7/9

The slope of a line perpendicular to the line:

We have already determined that the slope of the line is: -7/9

As we know that the slope of the perpendicular line is basically the negative reciprocal of the slope of the line.

Thus,  the slope of the perpendicular line will be:

\frac{-1}{\frac{-7}{9}}=\frac{9}{7}

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The Line integral is π/2.

Step-by-step explanation:

We have to find a funtion f such that its gradient is (ycos(xy), x(cos(xy)-ze^(yz), -ye^(yz)). In other words:

f_x = ycos(xy)

f_y = xcos(xy) - ze^{yz}

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we can find the value of f using integration over each separate, variable. For example, if we integrate ycos(x,y) over the x variable (assuming y and z as constants), we should obtain any function like f plus a function h(y,z). We will use the substitution method. We call u(x) = xy. The derivate of u (in respect to x) is y, hence

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\int {-ye^{yz}} \, dy = \int {-e^{u} \, dy} = -e^u +K = -e^{yz} + K(z)

Where, again, the constant of integration depends on Z.

As a result,

f(x,y,z) = cos(xy) - e^{yz} + K(z)

if we derivate f over z, we obtain

f_z(x,y,z) = -ye^{yz} + d/dz K(z)

That should be equal to -ye^(yz), hence the derivate of K(z) is 0 and, as a consecuence, K can be any constant. We can take K = 0. We obtain, therefore, that f(x,y,z) = cos(xy) - e^(yz)

The endpoints of the curve are r(0) = (0,0,1) and r(1) = (1,π/2,0). FOr the Fundamental Theorem for Line integrals, the integral of the gradient of f over C is f(c(1)) - f(c(0)) = f((0,0,1)) - f((1,π/2,0)) = (cos(0)-0e^(0))-(cos(π/2)-π/2e⁰) = 0-(-π/2) = π/2.

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