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Featured 2 months ago

Computing the rational number with fractional exponent

example of:

The rational number above

Prime Factorization :

Then we apply the power of a power with base

Featured 2 months ago

An equation to represent the perimeter in terms of its width is:

Let the width of the rectangle be

Let the length of the rectangle be

If the length (

The formula for perimeter of a rectangle is:

Substituting

Substituting

Sustituting

Featured 2 months ago

The legs are of length

**Method 1 - Familiar triangles**

The first few right angled triangles with an odd length side are:

#3, 4, 5#

#5, 12, 13#

#7, 24, 25#

Notice that

#15, 36, 39#

i.e.

Twice

**Method 2 - Pythagoras formula and a little algebra**

If the smaller leg is of length

#39 = sqrt(x^2 + (2x+6)^2)#

#color(white)(39) = sqrt(5x^2+24x+36)#

Square both ends to get:

#1521 = 5x^2+24x+36#

Subtract

#0 = 5x^2+24x-1485#

Multiply both sides by

#0 = 25x^2+120x-7425#

#color(white)(0) = (5x+12)^2-144-7425#

#color(white)(0) = (5x+12)^2-7569#

#color(white)(0) = (5x+12)^2-87^2#

#color(white)(0) = ((5x+12)-87)((5x+12)+87)#

#color(white)(0) = (5x-75)(5x+99)#

#color(white)(0) = 5(x-15)(5x+99)#

Hence

Discard the negative solution since we are seeking the length of the side of a triangle.

Hence the smallest leg is of length

Featured 2 months ago

Take the following steps to isolate and solve for

Subtract

Divide each side of the equation by

Featured 1 month ago

Square both sides:

#(sqrt(2x +48))^2 = x^2#

#2x + 48 = x^2#

#0 = x^2 - 2x - 48#

#0 = (x - 8)(x + 6)#

#x= 8 and -6#

Check, as extraneous solutions may have been introduced in the solving process.

#sqrt(2(8) + 48) =^? 8#

#sqrt(64) = 8" "color(green)(âˆš)#

AND

#sqrt(2(-6) + 48) =^? -6#

#sqrt(36) != -6" "color(red)(xx)#

Hopefully this helps!

Featured 3 weeks ago

â€¢Rectangle A: 6 by 7

â€¢Rectangle B: 7 by 3

The area of a rectangle is given by

The area of rectangle A is

The area of rectangle B is

We are given that the area of rectangle A is twice the area of rectangle B. Therefore, we can write the following equation.

A negative answer for

Therefore, the rectangles have the following measurements:

â€¢Rectangle A: 6 by 7

â€¢Rectangle B: 7 by 3

As you can see, rectangle A's area is twice the area of rectangle B, just as the problem indicated.

Hopefully this helps!

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