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kati45 [8]
3 years ago
5

43. Write a power represented with a positive base and a positive exponent

Mathematics
2 answers:
Levart [38]3 years ago
6 0
43. 5^3
44. Vcube =x^3, V=(1/3)^3 =1/27 (if the side of the cube is (1/3)
damaskus [11]3 years ago
6 0

Answer:

#43) 9^(1/2); #44) V = (1/3)^3 = 1/27

Step-by-step explanation:

#43) The only ways to make a power have a smaller value than the base is to use either a negative exponent (which we cannot do in this problem) or an exponent between 0 and 1 (a fraction or a decimal).

Finding a fractional power of a number is the same as taking that root of the number; for example, finding the 1/2 power of a number is the same as taking the square root; finding the 1/3 power of a number is the same as taking the cubed root; etc.

If we use 9 as the base (positive number) and 1/2 as the exponent (again a positive number), we would have 9^(1/2) = √9 = 3.

#44) To find the volume of any prism, find the area of the base and multiply it by the height.

For a cube, the base is a square; the area of a square is given by A = s², where s is the side length of the square.

Since the height of the cube will be the same as the length of the square (all sides are congruent in a cube), this makes the volume V = s²(s) = s³.

Since we know the side length is 1/3, this makes the volume

V = 1/3³ = 1/3(1/3)(1/3) = 1/9(1/3) = 1/27

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What is the minimum value of the objective function, C with given constraints? C=5x+3y {⎨x+3y≤9 {5x+2y≤20 {x≥1 {y≥2
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Answer:

The answer is below

Step-by-step explanation:

Plotting the following constraints using the online geogebra graphing tool:

x + 3y ≤ 9         (1)

5x + 2y ≤ 20    (2)

x≥1 and y≥2      (3)

From the graph plot, the solution to the constraint is A(1, 2), B(1, 2.67) and C(3, 2).

We need to minimize the objective function C = 5x + 3y. Therefore:

At point A(1, 2): C = 5(1) + 3(2) = 11

At point B(1, 2.67): C = 5(1) + 3(2.67) = 13

At point C(3, 2): C = 5(3) + 3(2) = 21

Therefore the minimum value of the objective function C = 5x + 3y is at point A(1, 2) which gives a minimum value of 11.

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