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morpeh [17]
3 years ago
6

Which ordered pairs in the form (x, y) are solutions to the equation (x + 6)y = 9? Choose all answers that are correct.  A.(−5,

9)  B.(−3, 3)  C.(3, 1)  D.(7, 9)
Mathematics
1 answer:
valentinak56 [21]3 years ago
4 0
<span>So we need to find the solutions to the equation (x+6)y=9 so we can arrange them in pair (x,y). We simply plug in the numbers into the equation. First pair (-5,9), we get (-5+6)9=9 so 1=1 and that is true, so first pair works. Second pair: (-3+6)3=9 we get (3)3=9 and 9=9 so second pair also works. Third pair: (3+6)1=9 we get 9=9, also works. Fourth pair: (7+6)9=9 we get 13*9=9 and 117=9 which is false so the fourth pair can't be a solution for the equation. The correct answers are A, B and C.</span>
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Find the solution of 3 times the square root of the quantity of x plus 5 equals negative 9, and determine if it is an extraneous
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Use the Fundamental Theorem for Line Integrals to find Z C y cos(xy)dx + (x cos(xy) − zeyz)dy − yeyzdz, where C is the curve giv
Harrizon [31]

Answer:

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}

f_z = -ye^{yz}

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

\int{ycos(xy)} \, dx = \int cos(u) \, du = sen(u) + C = sen(xy) + C(y,z)  

(Remember that c is treated like a constant just for the x-variable).

This means that f(x,y,z) = sen(x,y)+C(y,z). The derivate of f respect to the y-variable is xcos(xy) + d/dy (C(y,z)) = xcos(x,y) - ye^{yz}. Then, the derivate of C respect to y is -ze^{yz}. To obtain C, we can integrate that expression over the y-variable using again the substitution method, this time calling u(y) = yz, and du = zdy.

\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.

3 0
4 years ago
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