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Answer:

See a solution process below:

Explanation:

First, we can use this rule to combine the radicals within the parenthesis:

#root(n)(color(red)(a)) * root(n)(color(blue)(b)) = root(n)(color(red)(a) * color(blue)(b))#

#(root(4)(color(red)(x^3)) * root(4)(color(blue)(x^5)))^-2 = (root(4)(color(red)(x^3) * color(blue)(x^5)))^-2#

Next, use this rule for exponents to combine the terms within the radical:

#x^color(red)(a) xx x^color(blue)(b) = x^(color(red)(a) + color(blue)(b))#

#(root(4)(color(red)(x^3) * color(blue)(x^5)))^-2 = (root(4)(x^(color(red)(3)+color(blue)(5))))^-2 = (root(4)(x^8))^-2#

Then, we can use this rule to rewrite the radical into an exponent:

#root(color(red)(n))(x) = x^(1/color(red)(n))#

#(root(color(red)(4))(x^8))^-2 = ((x^8)^(1/color(red)(4)))^-2#

Next, we can use this rule to simplify the inner exponents:

#(x^color(red)(a))^color(blue)(b) = x^(color(red)(a) xx color(blue)(b))#

#((x^color(red)(8))^color(blue)(1/4))^-2 = (x^(color(red)(8) xx color(blue)(1/4)))^-2 = (x^2)^-2#

We can use the same rule to reduce the outer exponents:

#(x^color(red)(2))^color(blue)(-2) = x^(color(red)(2) xx color(blue)(-2)) = x^-4#

We can now use this rule to eliminate the negative exponent:

#x^color(red)(a) = 1/x^color(red)(-a)#

#x^color(red)(-4) = 1/x^color(red)(- -4) = 1/x^4#

Answer:

See a solution process belowL

Explanation:

First, write this equation in standard form by subtracting #color(red)(70z)# and adding #color(blue)(147)# to each side of the equation:

#7z^2 - color(red)(70z) + color(blue)(147) = 70z - color(red)(70z) - 147 + color(blue)(147)#

#7z^2 - 70z + 147 = 0 - 0#

#7z^2 - 70z + 147 = 0#

Next, divide each side of the equation by #color(red)(7)# to reduce the coefficients:

#(7z^2 - 70z + 147)/color(red)(7) = 0/color(red)(7)#

#(7z^2)/color(red)(7) - (70z)/color(red)(7) + 147/color(red)(7) = 0#

#1z^2 - 10z + 21 = 0#

#z^2 - 10z + 21 = 0#

Then we can factor the right side of the equation as:

#(z - 3)(z - 7) = 0#

Now, solve each term on the right side of the equation for #0#:

Solution 1:

#z - 3 = 0#

#z - 3 + color(red)(3) = 0 + color(red)(3)#

#z - 0 = 3#

#z = 3#

Solution 2:

#z - 7 = 0#

#z - 7 + color(red)(7) = 0 + color(red)(7)#

#z - 0 = 7#

#z = 7#

**The Solutions Are: #z = 3# and #z = 7#

We could also use the quadratic formula to solve this problem. The quadratic formula states:

For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))#

Substituting:

#color(red)(1)# for #color(red)(a)#

#color(blue)(-10)# for #color(blue)(b)#

#color(green)(21)# for #color(green)(c)# gives:

#z = (-color(blue)((-10)) +- sqrt(color(blue)((-10))^2 - (4 * color(red)(1) * color(green)(21))))/(2 * color(red)(1))#

#z = (10 +- sqrt(color(blue)(100 - 84)))/2#

#z = (10 - sqrt(16))/2# and #z = (10 + sqrt(16))/2#

#z = (10 - 4)/2# and #z = (10 + 4)/2#

#z = 6/2# and #z = 14/2#

#z = 3# and #z = 7#

Answer:

#1/5^(n+7)#

Explanation:

In order to simplify exponents, make sure that the bases are all the same. Change numbers to the product of their prime factors.

#(color(red)(25)^(2n-4))/(5^(3n+1) xx 5^(2n-3) xx 5) = (color(red)((5^2))^(2n-4))/(5^(3n+1) xx 5^(2n-3) xx 5)#

#=(color(red)(5)^(4n-8))/(5^(3n+1) xx 5^(2n-3) xx 5^1)#

Add the indices of like bases when multiplying

#(5^(4n-8))/(5^(5n-1)#

Method 1

Subtract the indices when dividing

#= 1/5^(5n-4n -1-(-8))#

#=1/5^(n+7)#

Method 2

#x^m = 1/(x^-m)#

#(5^(4n-8))/(5^(5n-1)#

#=1/(5^(5n-1) xx 5^(-4n+8))#

#=1/5^(n+7)#

Give the answer without negative or zero indices.

Answer:

no real roots; #Delta<0#

Explanation:

#-2x^2-x-1=0# is already in #ax^2+bx+c# form, so the #a#,#b# and #c# -values can be used.

#a=-2#
#b=-1#
#c=-1#

#Delta=b^2-4ac#
#=(-1)^2 - (4*(-2*-1))#
#=1-8#
#-7#

#-7<0 therefore Delta<0#

hence, #-2x^2-x-1=0# has no real roots.

this can also be seen by graphing #-2x^2-x-1=0#:

https://www.desmos.com/calculator

this parabola does not meet the #x#-axis on a graph for real numbers, so there are no (real) roots.

Answer:

#3/5 xx (-1/4-1/6) -:(-7/3+5/4)=color(blue)(3/13#

Explanation:

Simplify:

#3/5 xx (-1/4-1/6) -:(-7/3+5/4)#

Follow the order of operations: parentheses, exponents, multiplication and division left to right, addition and subtraction left to right.

Simplify the parentheses first.

#(-1/4-1/6)# must have a common denominator. The least common denominator (LCD) can be found by listing the multiples of #4# and #6#.

#4:##4,8,color(red)12,16...#

#6:##6,color(red)12...#

LCD#=##color(red)12#

Multiply both fractions by a fraction equal to #1#, so that each will have the same denominator. For example, #5/5=1#. This way, the values don't change.

#(-1/4xxcolor(teal)3/color(teal)3-1/6xxcolor(magenta)2/color(magenta)2)#

Simplify.

#(-3/12-2/12)=(-5/12)#

Return #(-5/12)# to the original expression.

#3/5 xx (-5/12) -:(-7/3+5/4)#

Simplify #(-7/3+5/4)#.

Follow the same procedure as with #(-1/4-1/6)#.

LCD#=##12#

#(-7/3xxcolor(green)4/color(green)4+5/4xxcolor(red)3/color(red)3)#

Simplify.

#(-28/12+15/12)=-13/12#

Return #(-13/12)# to the original equation.

#3/5 xx (-5/12) -:(-13/12)#

Multiply #3/5# and #-5/12#.

#-15/60-:(-13/12)#

Divide #-15/60# and #-13/12#.

When dividing by a fraction, invert the fraction and multiply.

#-15/60xx-12/13#

Simplify.

#180/780#

Reduce by dividing the numerator and denominator by #60#.

#(180-:60)/(780-:60)=3/13#

Answer:

See below.

Explanation:

#y = log_2 3 hArr 2^y = 3# If #y# is rational then #y = n/m# with #{n,m} in ZZ# and # m ne 0# or

#2^(n/m) = 3 rArr 2^n = 3^m#

This last equality is an absurd because #3^m# is odd and #2^n# is even.

Concluding #y = log_2 3# is not rational, being then irrational.

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