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
Answer:
Step-by-step explanation:
6.0 , 6.0 , 6.1 , 6.2 , 6.3 , 6.15 , 6.23
hope that help mark me as a brill
<h2>
Answer:</h2>
The values of x for which the given vectors are basis for R³ is:

<h2>
Step-by-step explanation:</h2>
We know that for a set of vectors are linearly independent if the matrix formed by these set of vectors is non-singular i.e. the determinant of the matrix formed by these vectors is non-zero.
We are given three vectors as:
(-1,0,-1), (2,1,2), (1,1, x)
The matrix formed by these vectors is:
![\left[\begin{array}{ccc}-1&2&1\\0&1&1\\-1&2&x\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%262%261%5C%5C0%261%261%5C%5C-1%262%26x%5Cend%7Barray%7D%5Cright%5D)
Now, the determinant of this matrix is:

Hence,

This is a trick question. They eat 1/3 of the gallons given. Multiply
1/3 * 1/3 = 1/9 gallons or 0.111 gallons
47/9+110/15=705/135+990/135=1695/135=12 15/27