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Alexus L.

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Linear Algebra

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Question:

Prove that if A is an nxn matrix, then det(A^-1) =(detA)^-1

Alexus L.

Answer:

AA^-1 = I and det I =1 using the Theorem det(AB)=det(A)det(B) we get det(A)det(A^-1) = det(AA^-1) = det(I) = 1 Therefore det(A^-1) = 1/(detA) = (detA)^-1

Chemistry

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Question:

The number 0.0043015 has how many significant figures? Also, what is its scientific notation?

Alexus L.

Answer:

The number has 5 significant figures seeing as though leading zeros do not matter. The zero in the middle is included. The scientific notation for this number would be 4.3015 x 10^3. In order to keep the same number of significant figures the numbers behind the 3 must be included. If it is rounded to 4.3 x 10^3 the number of significant figures will go down to 2.

Calculus

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Question:

A flat piece of ice is shaped like a square. When the ice melts the area of the sheet is decreasing at a rate of 0.25 m^2/sec. What is the decreasing rate of the length when the area of the ice is 25 m^2 ?

Alexus L.

Answer:

The area of the square with respect to time can be defined as : A(t)= lw where l=length of the square and w= width of the square. Since we are dealing with a square we will simplify our equation to A(t)= l^2 since the our sides are all the same size. From the problem we can see that the rate of decrease for our are (dA/dt) = -0.25 and we are asked to find (dl/dt) when A(t)=25. Next we must differentiate our equation with respect to t. Our new equation then becomes dA/dt = (2)(l)(dl/dt) Finally, we must go back to our original equation and plug in to find our length since we were given the area at that specific time. 25= l^2 so, l=5 The rate of decrease for our length is then (-0.25) = (2)(5)(dl/dt) so, dl/dt = -0.025.

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