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PIT_PIT [208]
3 years ago
14

Tamika has a hard rubber ball whose circumference measures 13 inches. she wants to box it for a gift but can only find cube-shap

ed boxes of sides 3 inches, 4 inches, 5 inches, or 6 inches. which of these is the smallest box that the ball will fit into with the top on? 1.
Mathematics
1 answer:
Ostrovityanka [42]3 years ago
3 0
The rubber ball, in this case, can be considered a circle with a circumference of 13 inches. So, we can find the diameter of the rubber ball using the formula C=\pi d or d=\frac{C}{\pi }. That is 
    d=\frac{C}{\pi }=\frac{13\:inches}{\pi }=\frac{13}{\pi }=4.14\:inches

Therefore, the diameter is greater than 4 inches and less than 5 inches. That means the smallest box that the ball will fit into with the top on is the 5 inches box. 


     
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Find the directional derivative of f(x,y,z)=z3−x2yf(x,y,z)=z3−x2y at the point (−5,5,2)(−5,5,2) in the direction of the vector v
olga_2 [115]

We are given

f=z^3 -x^2y

Firstly, we can find gradient

so, we will find partial derivatives

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f_y=-x^2

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now, we can plug point (-5,5,2)

f_x=-2*-5*5=50

f_y=-(-5)^2=-25

f_z=3(2)^2=12

so, gradient will be

gradf=(50,-25,12)

now, we are given that

it is in direction of v=⟨−3,2,−4⟩

so, we will find it's unit vector

|v|=\sqrt{(-3)^2+(2)^2+(-4)^2}

|v|=\sqrt{29}

now, we can find unit vector

v'=(\frac{-3}{\sqrt{29} } , \frac{2}{\sqrt{29} } , \frac{-4}{\sqrt{29} })

now, we can find dot product to find direction of the vector

dir=(gradf) \cdot (v')

now, we can plug values

dir=(50,-25,12) \cdot (\frac{-3}{\sqrt{29} } , \frac{2}{\sqrt{29} } , \frac{-4}{\sqrt{29} })

dir=(-\frac{150}{\sqrt{29} } - \frac{50}{\sqrt{29} } - \frac{48}{\sqrt{29} })

dir=-\frac{248\sqrt{29}}{29}.............Answer



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Step-by-step explanation:

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