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vovikov84 [41]
3 years ago
11

why do whole numbers raised to an exponent get greater while fractions raised to an exponent get smaller?

Mathematics
2 answers:
ehidna [41]3 years ago
8 0
Any number above 1 gets greater, below 1 smaller (when above 0), while 1 itself remains the same. Negative numbers are more unpredictable.
8_murik_8 [283]3 years ago
8 0

Answer:

Because whole numbers are greater than 1 and any number greater than 1 multiplied by it self several times result in a greater number, but proper fractions are numbers less than 1, and a number less than 1 multiplied by it self several times result if a smaller number.

Explanation:

Whole numbers are raised to an exponent , means that such number is multiplied by itself (as many times and the exponent is).

Whole numbers are 1, 2, 3, 4, ....

Number 1 is a special case. When 1 is multiplied by it self the result is 1.

When other whole number (2, 3, 4, 5, ...) is multiplied by it self the result is a greater number. For example, 2 × 2 = 4, 3 × 3 = 9, and so on.

On the other hand, proper fractions, which are those whose numerator is less than the denominator, are less than one, and any number less than 1 multiplied by it self results in a smaller number.

For example: 1/2 × 1/2 = 1/4 or 0.5 × 0.5 = 0.25. This is always true for any proper fraction (a number less than 1).


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Answer:

No solution

Explanation

−x^2+5x−10=0

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you plug it in as shown below

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3 years ago
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I would like to create a rectangular vegetable patch. The fencing for the east and west sides costs $4 per foot, and the fencing
Anestetic [448]
Let the lengths of the east and west sides be x and the lengths of the north and south sides be y.  the dimensions you want are therefore x and y.

The cost of the east and west fencing is $4*2*x; the cost of the north and south fencing is $2*2*y.  We have to put in that "2" because there are 2 sides that run from east to west and 2 sides that run from north to south.



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Now we are to maximize the area of the vegetable patch, subject to the constraint that 2x + y = 32.  The formula for area is A = L * W.  Solving 2x + y = 32 for y, we get y = -2x + 32.

We can now eliminate y.  The area of the patch is (x)(-2x+32) = A.  We want to maximize A.

If you're in algebra, find the x-coordinate of the vertex of this quadratic equation.  Remember the formula x = -b/(2a)?  Once you have calculated this x, subst. your value into the formula for y:  y= -2x + 32.

Now multiply together your x and y values to obtain the max area of the patch.


If you're in calculus, differentiate A = x(-2x+32) with respect to x and set the derivative equal to zero.  This approach should give you the same x value as before; the corresponding y value will be the same;  y=-2x+32.

Multiply x and y together.  That'll give you the maximum possible area of the garden patch.
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3 years ago
What is the vaule of 12(3x+5x),when x=7
german
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