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const2013 [10]
3 years ago
13

A line has a slope of 5 and passes through the point (1, 0). What is its equation in

Mathematics
2 answers:
serg [7]3 years ago
8 0

Answer:

y = 5x - 5

Step-by-step explanation:

y = mx + b

m = 5

b = ?

Plug in the coordinates

0 = 5(1) + b

0 = 5 + b

-5 = b

Finally:

y = 5x - 5

mars1129 [50]3 years ago
4 0

Answer:

y=5x-5

Step-by-step explanation:

find y-intercept:

0=1(5)+b

b=-5

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\left[\begin{matrix} 0 & 9 \\ 9 & 0\end{matrix}\right]

Note that the first determinant is 0, and the second determinant is -9. THis tell us that the point  is a saddle point, hence not a minimum nor maximum.  

Since the function is continous and the region R is closed and bound (hence compact) the maximum and minimum must be attained on the boundaries of R. REcall that when -2\leq x \leq x and y=0 we have that F(x,0) = 0. So, we want to pay attention to the critical values over the circle, restricting that the values of y must be positive. To do so, consider the following function

H(x,y, \lambda) = 9xy - \lambda(x^2+y^2-4) which consists of the original function and a function that describes the restriction (the circle x^2+y^2=4), we want that the gradient of H is 0.

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\frac{dH}{dx} = 9y-2\lambda x =0

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F(\sqrt[]{2},\sqrt[]{2}) = 9\cdot 2 =18

as for the second point the value of the function is

F(-\sqrt[]{2},\sqrt[]{2}) = 9\cdot -2 =-18.

Then, the point (\sqrt[]{2},\sqrt[]{2}) is a maximum and the point (-(\sqrt[]{2},\sqrt[]{2}) is a minimum.

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3 years ago
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