Find the surface area of the sphere with the given dimension. Leave your answer in terms of π. Sphere with radius of 60 m.

1 answer:

6 0

The surface area of a sphere is given by $A=4 \pi r^{2}$ where r represents the radius.

As the radius is given as 60 meters we replace r with 60 in the formula above to get the surface area.

$A=4 \pi ( 60^{2} )=14400 \pi$

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What is m<4? I tryed to do the work but I just can't get it can someone help me plz

Mars2501 [29]

M<6 is 60
Since AC and BD are parallel, <6 and <4 are supplementary.
supplementary means = 180 degrees
Since m<6 is 60 degrees, subtract 60 from 180 to get <4
180 - 60 = 120

M<4 is 120

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A company manufactures running shoes and basketball shoes. The total revenue (in thousands of dollars) from x1 units of running

Alborosie

Answer:

$x_1 =2 , x_2=7$

Step-by-step explanation:

Consider the revenue function given by $R(x_1,x_2) = -5x_1^2-8x_2^2 -2x_1x_2+34x_1+116x_2$ . We want to find the values of each of the variables such that the gradient( i.e the first partial derivatives of the function) is 0. Then, we have the following (the explicit calculations of both derivatives are omitted).

$\frac{dR}{dx_1} = -10x_1-2x_2+34 =0$

$\frac{dR}{dx_2} = -16x_2-2x_1+116 =0$

From the first equation, we get, $x_2 = \frac{-10x_1+34}{2}$ .If we replace that in the second equation, we get

$-16\frac{-10x_1+34}{2} -2x_1+116=0= 80x_1-2x_1+116-272= 78x_1-156$

From where we get that $x_1 = \frac{156}{78}=2$ . If we replace that in the first equation, we get

$x_2 = \frac{-10\cdot 2 +34}{2}=\frac{14}{2} = 7$

So, the critical point is $(x_1,x_2) = (2,7)$ . We must check that it is a maximum. To do so, we will use the Hessian criteria. To do so, we must calculate the second derivatives and the crossed derivatives and check if the criteria is fulfilled in order for it to be a maximum. We get that

$\frac{d^2R}{dx_1dx_2}= -2 = \frac{d^2R}{dx_2dx_1}$

$\frac{d^2R}{dx_{1}^2}=-10, \frac{d^2R}{dx_{2}^2}=-16$

We have the following matrix,

$\left[\begin{matrix} -10 & -2 \\ -2 & -16\end{matrix}\right]$ .

Recall that the Hessian criteria says that, for the point to be a maximum, the determinant of the whole matrix should be positive and the element of the matrix that is in the upper left corner should be negative. Note that the determinant of the matrix is $(-10)\cdot (-16) - (-2)(-2) = 156>0$ and that -10<0. Hence, the criteria is fulfilled and the critical point is a maximum