The tangent line to a curve is the one that coincides with the curve at a point and with the same derivative, that is, the same degree of variation.
We have then:
y = 5x-x²
Deriving:
y '= 5-2x
In point (1, 4)
The slope is:
y (1) '= 5-2 * (1)
y (1) '= 3
The equation of the line will be:
y-f (a) = f '(a) (x-a)
We have then:
y-4 = 3 (x-1)
Rewriting:
y = 3x-3 + 4
y = 3x + 1
Answer:
the tangent line to the parabola at the point (1, 4) is
y = 3x + 1
the slope m is
m = 3
<span>Simplifying
(7 + -2x)(11 + -2x)(x) = 0
Reorder the terms for easier multiplication:
x(7 + -2x)(11 + -2x) = 0
Multiply (7 + -2x) * (11 + -2x)
x(7(11 + -2x) + -2x * (11 + -2x)) = 0
x((11 * 7 + -2x * 7) + -2x * (11 + -2x)) = 0
x((77 + -14x) + -2x * (11 + -2x)) = 0
x(77 + -14x + (11 * -2x + -2x * -2x)) = 0
x(77 + -14x + (-22x + 4x2)) = 0
Combine like terms: -14x + -22x = -36x
x(77 + -36x + 4x2) = 0
(77 * x + -36x * x + 4x2 * x) = 0
(77x + -36x2 + 4x3) = 0
Solving
77x + -36x2 + 4x3 = 0
Solving for variable 'x'.
Factor out the Greatest Common Factor (GCF), 'x'.
x(77 + -36x + 4x2) = 0
Factor a trinomial.
x((7 + -2x)(11 + -2x)) = 0
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Use a ruler for this problem. Take a picture of the measurement & then I could help you.
I think the answer is 13 x 6(2) = 156? Sorry if I’m wrong.
Answer:
last one is the square matrix
