A line passes through the point (-6, -2) and has a slope of -5/2
2 answers:
Answer:
y = (-5/2)x - 17
Step-by-step explanation:
<h2>Slope Intercept Formula:</h2>y = mx + b
m = slope
b = y-intercept
<h2>Plugin Information:</h2>
y = mx + b
Slope = -5/2
y = (-5/2)x + b Input x and y values from given point
-2 = (-5/2)-6 + b
<h2>Simplify, Solve for b:</h2>-2 = (-5 * -6)/(2) + b
-2 = 30/2 + b
-2 = 15 + b
b = -2 - 15
b = -17
<h2 /><h2>Rewrite Equation:</h2>y = mx + b
m = -5/2
b = -17
y = (-5/2)x - 17
-Chetan K
-Chetan K
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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:
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
(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.
Where, again, the constant of integration depends on Z.
As a result,
if we derivate f over z, we obtain
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.