Answer:
There is no point of the form (-1, y) on the curve where the tangent is horizontal
Step-by-step explanation:
Notice that when x = - 1. then dy/dx becomes:
dy/dx= (y+2) / (2y+1)
therefore, to request that the tangent is horizontal we ask for the y values that make dy/dx equal to ZERO:
0 = ( y + 2) / (2 y + 1)
And we obtain y = -2 as the answer.
But if we try the point (-1, -2) in the original equation, we find that it DOESN'T belong to the curve because it doesn't satisfy the equation as shown below:
(-1)^2 + (-2)^2 - (-1)*(-2) - 5 = 1 + 4 + 2 - 5 = 2 (instead of zero)
Then, we conclude that there is no horizontal tangent to the curve for x = -1.
Answer:
The value of x is 12
Step-by-step explanation:
To find the value of x, we need to note that the interior angles are equal to 180. We also know that angle R is equal to 180 - (8 + 6x). So we can add all of this together and set equal to 180.
180 - (8 + 6x) + 4x + 2 + 30 = 180
180 - 8 - 6x + 4x + 2 + 30 = 180
-2x + 24 = 0
-2x = -24
x = 12
Answer:
Step-by-step explanation:
So 8 times what is 5?
8 times x=5
x=0.625
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a+b+c=0
[(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc]
[a^2+b^2+c^2+2ab+2ac+2bc=0]
[a^2+b^2+c^2=-(2ab+2ac+2bc)]
[a^2+b^2+c^2=-2(ab+ac+bc)] (i)
also
[a=-b-c]
[a^2=-ab-ac] (ii)
[-c=a+b]
[-bc=ab+b^2] (iii)
adding (ii) and (iii) ,we have
[a^2-bc=b^2-ac] (iv)
devide (i) by (iv)
[(a^2+b^2+c^2)/(a^2-bc)=(-2(ab+bc+ca))/(b^2-ac)]
