Answer:
-4/5
Step-by-step explanation:
To find the slope of the tangent to the equation at any point we must differentiate the equation.
x^3y+y^2-x^2=5
3x^2y+x^3y'+2yy'-2x=0
Gather terms with y' on one side and terms without on opposing side.
x^3y'+2yy'=2x-3x^2y
Factor left side
y'(x^3+2y)=2x-3x^2y
Divide both sides by (x^3+2y)
y'=(2x-3x^2y)/(x^3+2y)
y' is the slope any tangent to the given equation at point (x,y).
Plug in (2,1):
y'=(2(2)-3(2)^2(1))/((2)^3+2(1))
Simplify:
y'=(4-12)/(8+2)
y'=-8/10
y'=-4/5
Answer:
u=92
Step-by-step explanation:
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Answer:
The desired equation is y = (-8/3)x + 26/3.
Step-by-step explanation:
Moving from (1,6) to (4, -2) involves an increase of 3 in x and a decrease of 8 in y. Thus, the slope of the line thru these two points is m = rise / run = -8/3.
Using the slope-intercept form of the eq'n of a straight line and inserting the data given (slope = m = -8/3, x = 4, y = -2), we get:
y = mx + b => -2 = (-8/3)(4) + b, or -2 = -32/3 + b
Multiply all terms by 3 to clear out the fraction:
-6 = -32 + 3b.
Then 26 = 3b, and b = 26/3.
The desired equation is y = (-8/3)x + 26/3.
Answer:52
Step-by-step explanation:
decreased by -12 is basically adding 12. so 40+12=52
The correct answers are, <u>C and D</u>
