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goldenfox [79]
3 years ago
11

Which shows the dimensions of two rectangular prisms that have volumes of 72 cm3 but different surface areas? A. 6 cm by 3 cm by

4 cm; 12 cm by 2 cm by 3 cm B. 2 cm by 4 cm by 9 cm; 9 cm by 4 cm by 2 cm C. 3 cm by 3 cm by 8 cm; 2 cm by 6 cm by 8 cm D. 6 cm by 3 cm by 4 cm; 9 cm by 4 cm by 3 cm
Mathematics
2 answers:
Nuetrik [128]3 years ago
4 0

Answer:

A

Step-by-step explanation:

A, is the answer

Nookie1986 [14]3 years ago
3 0
To solve, lets find the volumes of all of the options...

V=l*w*h

A. 
6*3*4=72cm³
12*2*3=72cm³

B.
2*4*9=72cm³
9*4*2=72cm³

C.
3*3*8=72cm³
2*6*8=96cm³

D.
6*3*4=72cm³
9*4*3=108cm³

We can conclude that C & D aren't the answer, since they contain prisms that don't have a volume of 72cm³.

Now lets solve for the surface area of A and B...

A. 
sA=2(wh+lw+lh)=2(6*3+4*6+4*3)=2(18+24+12)=2(54)=108cm²
sA=2(wh+lw+lh)=2(12*2+3*12+3*2)=2(24+36+6)=2(56)=112cm²


B.
sA=2(wh+lw+lh)=2(2*4+9*2+9*4)=2(8+18+36)=2(62)=124cm²
sA=2(wh+lw+lh)=2(9*4+2*9+2*4)=2(36+18+8)=2(62)=124cm²

A is the only option with both similar volumes of 72cm³ and different surface areas...

Answer=A
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Does there exist a di↵erentiable function g : [0, 1] R such that g'(x) = f(x) for all x 2 [0, 1]? Justify your answer
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No; Because g'(0) ≠ g'(1), i.e. 0≠2, then this function is not differentiable for g:[0,1]→R

Step-by-step explanation:

Assuming:  the function is f(x)=x^{2} in [0,1]

And rewriting it for the sake of clarity:

Does there exist a differentiable function g : [0, 1] →R such that g'(x) = f(x) for all g(x)=x² ∈ [0, 1]? Justify your answer

1) A function is considered to be differentiable if, and only if  both derivatives (right and left ones) do exist and have the same value. In this case, for the Domain [0,1]:

g'(0)=g'(1)

2) Examining it, the Domain for this set is smaller than the Real Set, since it is [0,1]

The limit to the left

g(x)=x^{2}\\g'(x)=2x\\ g'(0)=2(0) \Rightarrow g'(0)=0

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g'(x)=f(x) then g'(0)=f(0) and g'(1)=f(1)

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Because this is the same as to calculate the limit from the left and right side, of g(x).

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This is what the Bilateral Theorem says:

\lim_{x\rightarrow c^{-}}f(x)=L\Leftrightarrow \lim_{x\rightarrow c^{+}}f(x)=L\:and\:\lim_{x\rightarrow c^{-}}f(x)=L

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4 years ago
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11111nata11111 [884]

Answer:

A) The point( -1 , 5)  is satisfies the  given graph  g(x) = (\frac{1}{5} )^{x}

B) The point( 3 , 1/125)  is satisfies the  given graph  g(x) = (\frac{1}{5} )^{x}

Step-by-step explanation:

<u><em>Explanation:-</em></u>

Given graph  

               y =      g(x) = (\frac{1}{5} )^{x}

Put the point ( -1 , 5)

Put y =5 and x =-1

            5 = (\frac{1}{5} )^{-1} = 5

The point( -1 , 5)  is satisfies the  given graph

ii)

Given graph  

               y =      g(x) = (\frac{1}{5} )^{x}

             \frac{1}{125}  = (\frac{1}{5} )^{3} = \frac{1}{125}

The point( 3 , 1/125)  is satisfies the  given graph  g(x) = (\frac{1}{5} )^{x}

                 

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