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Evaluate the following?

Mia
Mia
Featured 2 months ago

Answer:

#7/11#

Explanation:

Computing the rational number with fractional exponent #(u/v)^(1/6)=u^(1/6)/v^(1/6)# is determined by prime factorizing the numerator and denominator .

example of: #sqrtx = x^(1/2#:

#sqrt((4/9)) = (4/9)^(color(blue)(1/2))=(2^2)^color(blue)(1/2)/(3^2)^color(blue)(1/2)=2/3#

The rational number above #(117649/1771561)^(1/6)=(117649)^(1/6)/(1771561)^(1/6)#

Prime Factorization :
#color(blue)(117649=7xx7xx7xx7xx7xx7=7^6)#
#color(brown)(1771561=11xx11xx11xx11xx11xx11=11^6)#

#(117649/1771561)^(1/6)=(117649)^(1/6)/(1771561)^(1/6)=(color(blue)(7^6))^(1/6)/(color(brown)(11^6))^(1/6)#

Then we apply the power of a power with base #a# :
#(a^m)^(1/n)=a^(m/n)#

#(117649/1771561)^(1/6)=(7^(6/6)/11^(6/6))=7/11#

Answer:

An equation to represent the perimeter in terms of its width is: #p = 4w + 14# and the length of the rectangle is #10# ft.

Explanation:

Let the width of the rectangle be #w#.

Let the length of the rectangle be #l#.

If the length (#l#) is 7 feet longer than the width, then the length can be written in terms of the width as:

#l = w + 7#

The formula for perimeter of a rectangle is:

#p = 2l + 2w# where #p# is the perimeter, #l# is the length and #w# is the width.

Substituting #w + 7# for #l# gives an equation to represent the perimeter in terms of its width:

#p = 2(w + 7) + 2w#

#p = 2w + 14 + 2w#

#p = 4w + 14#

Substituting #26# for #p# allows us to solve for #w#.

#26 = 4w + 14#

#26 - 14 = 4w + 14 - 14#

#12 = 4w#

#12/4 = 4w / 4#

#w = 3#

Sustituting #3# for #w# in the equation above, #l = w + 7# allows us to determine the length:

#l = 3 + 7#

#l = 10#

Answer:

The legs are of length #15# and #36#

Explanation:

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 #39 = 3 * 13#, so will a triangle with the following sides work:

#15, 36, 39#

i.e. #3# times larger than a #5, 12, 13# triangle ?

Twice #15# is #30#, plus #6# is #36# - Yes.

#color(white)()#
Method 2 - Pythagoras formula and a little algebra

If the smaller leg is of length #x#, then the larger leg is of length #2x+6# and the hypotenuse is:

#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 #1521# from both sides to get:

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

Multiply both sides by #5# to get:

#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 #x = 15# or #x = -99/5#

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

Hence the smallest leg is of length #15# and the other is #2*15+6 = 36#

Answer:

#x = 5#

Explanation:

Take the following steps to isolate and solve for #x# while keeping the equation balanced.

Subtract #9# from each side of the equation:

#6x + 9 - 9 = 39 - 9#

#6x + 0 = 30#

#6x = 30#

Divide each side of the equation by #6#

#(6x)/6 = 30/6#

#1x = 5#

#x = 5#

Answer:

#{8}#.

Explanation:

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)(√)#

AND

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

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

Hopefully this helps!

Answer:

•Rectangle A: 6 by 7
•Rectangle B: 7 by 3

Explanation:

The area of a rectangle is given by #color(red)(A = l * w)#.

The area of rectangle A is #6(10 - x) = 60 - 6x#

The area of rectangle B is #x(2x + 1) = 2x^2 + x#

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

#60 - 6x = 2(2x^2 + x)#

#60 - 6x = 4x^2 + 2x#

#0 = 4x^2 + 8x - 60#

#0 = 4(x^2 + 2x - 15)#

#0 = (x + 5)(x - 3)#

#x = -5 and 3#

A negative answer for #x# is impossible, since we're talking about geometric shapes.

Therefore, the rectangles have the following measurements:

•Rectangle A: 6 by 7
•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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