If the number in numeratorof a unit rate is 1,what does this indicate about the equivalent unit rates

1 answer:

6 0

Unit rate The rate for one unit of given quantity is called a unit rate. These unit rates are again considered as one quantity. Examples are km/h which is derived for instance given that the distance is 10km and the time to be there is 5 hours hence what is the velocity. Which is v=d/t henceafter, 2km/h. Another example is pressure which is force divided by the area. 1 N/m2 which is 5 N times 5 m2, the pressure is 1 N/m2

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If f(x) = 3x - 1 and g(x) = x + 2, find (f - g)(x).

Marina86 [1]

Answer:

$\boxed{\sf (f - g)(x) = 2x - 3}$

Given:

f(x) = 3x - 1

g(x) = x + 2

To Find:

(f - g)(x)

Step-by-step explanation:

$\sf (f -g)(x) = f(x) - g(x) \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =(3x - 1) - (x + 2) \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =3x - 1 - x - 2 \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =3x - x - 1 - 2 \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =(3x - x) - (1 + 2) \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =2x - (1 + 2) \\ \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =2x - 3$

6 0

3 years ago

EXAMPLE 5 Find the maximum value of the function f(x, y, z) = x + 2y + 9z on the curve of intersection of the plane x − y + z =

geniusboy [140]

The Lagrangian,

$L(x,y,z,\lambda,\mu)=x+2y+9z-\lambda(x-y+z-1)-\mu(x^2+y^2-1)$

has critical points where its partial derivatives vanish:

$L_x=1-\lambda-2\mu x=0$

$L_y=2+\lambda-2\mu y=0$

$L_z=9-\lambda=0$

$L_\lambda=x-y+z-1=0$

$L_\mu=x^2+y^2-1=0$

$L_z=0$ tells us $\lambda=9$ , so that

$L_x=0\implies-8-2\mu x=0\implies x=-\dfrac4\mu$

$L_y=0\implies11-2\mu y=0\implies y=\dfrac{11}{2\mu}$

Then with $L_\mu=0$ , we get

$x^2+y^2=\dfrac{16}{\mu^2}+\dfrac{121}{4\mu^2}=1\implies\mu=\pm\dfrac{\sqrt{185}}2$

and $L_\lambda=0$ tells us

$x-y+z=-\dfrac4\mu-\dfrac{11}{2\mu}+z=1\implies z=1+\dfrac{19}{2\mu}$

Then there are two critical points, $\left(\pm\frac8{\sqrt{185}},\mp\frac{11}{\sqrt{185}},1\pm\frac{19}{\sqrt{185}}\right)$ . The critical point with the negative $x$ -coordinates gives the maximum value, $9+\sqrt{185}$ .