Answer:
you dint put up the pictures so not sure
Step-by-step explanation:
What's the problem? There's no example...
Answer:
x=-2, y=5. (-2, 5).
Step-by-step explanation:
15x-4y=-50
3x-2y=-16
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15x-4y=-50
-5(3x-2y)=-5(-16)
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15x-4y=-50
-15x+10y=80
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6y=30
y=30/6
y=5
3x-2(5)=-16
3x-10=-16
3x=-16+10
3x=-6
x=-6/3
x=-2
Answer: The equations of the tangents that pass through
y = -x- 1 and
y = 11x -25
Step-by-step explanation: The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. So for our curve (the parabola) we have
y= x^2 +x
Differentiating wrt x we get:
dy/dx = 2x +1
Let P (a, B) be any generic point on the curve. Then the gradient of the tangent at P is given by:
