Sabrina would have to drive approximately 10 1/2 more hours.
Answer:
mm
kl l
Step-by-step explanation:
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:
The base is: ![3 \sqrt[3]{4}](https://tex.z-dn.net/?f=3%20%5Csqrt%5B3%5D%7B4%7D)
Step-by-step explanation:
Given
![f(x) = \frac{1}{4}(\sqrt[3]{108})^x](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B1%7D%7B4%7D%28%5Csqrt%5B3%5D%7B108%7D%29%5Ex)
Required
The base
Expand 108
![f(x) = \frac{1}{4}(\sqrt[3]{3^3 * 4})^x](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B1%7D%7B4%7D%28%5Csqrt%5B3%5D%7B3%5E3%20%2A%204%7D%29%5Ex)
Rewrite the exponent as:
Expand
Rewrite as:
![f(x) = \frac{1}{4}(3 \sqrt[3]{4})^x](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Cfrac%7B1%7D%7B4%7D%283%20%5Csqrt%5B3%5D%7B4%7D%29%5Ex)
An exponential function has the following form:
Where
By comparison:
![b =3 \sqrt[3]{4}](https://tex.z-dn.net/?f=b%20%3D3%20%5Csqrt%5B3%5D%7B4%7D)
So, the base is:
