1. -4x^3 times 2y^-2 times 5y^5 times x^-8
2. (n^8)^3
3. (t^-2)^5
4. (3q^2)^4
2007-10-12 12:11:34 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
I am GUESSING you mean 'simplify' the expressions, since there is nothing to solve!
1. -4x^3 * 2y^-2 * 5y^5 * x^-8, rewrite as
-4*2*5 * x^3 * x^-8* y^-2 * y^5 * , ie, get numbers together, x's together, y's together; simplify
-40 * x^-5* y^3 * , by rules of exponents. Done.
2. (n^8)^3. Write as
n^8 * n^8 * n^ 8. How many n's do you have?
= n^24, by rules of exponents.
Know these rules, you can do the rest.
2007-10-12 12:21:05 · answer #1 · answered by Anonymous · 0⤊ 0⤋
1. -4x^3 times 2y^-2 times 5y^5 times x^-8
= -4(2)(5)y^3/x^5 = -40y^3/x^5
2. (n^8)^3 = n^24
3. (t^-2)^5 = t^-10 = 1/t^10
4. (3q^2)^4 = 81q^8
2007-10-12 12:18:40 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
-4x^3*2y^2*5y^5*x^-8 = (-4x^3*x^-8)*(2y^2*5y^5)
>>>>>>>>>>>>>>>> = -4x^(3-8)*(2*5)y^(2+5)
>>>>>>>>>>>>>>>> = (-4x^-5)(10y^7)
(ax^m)^n = (a^mn)(x^mn) => (n^8)^3 = n^24
=> (t^-2)^5 = t^-10
=> (3q^2)^4 = (3^4)(q^2)^4
>>>>>>>>> = 81q^8
2007-10-12 12:25:48 · answer #3 · answered by Helen B 5 · 0⤊ 0⤋
Use the formula x^m * x^n = x^(m+n)
and (x^n)^m = x^(n*m)
1. -40 x^(-5) * y ^3
2. n^24
3. t^(-10)
4. 81q^8
2007-10-12 12:18:09 · answer #4 · answered by norman 7 · 0⤊ 0⤋
It's Friday, put the books up and go have fun
2007-10-12 12:17:23 · answer #5 · answered by Big Blue Wolverine 3 · 0⤊ 0⤋
those don't even make sense.
Are these equations? or have you been given values for the variables?
2007-10-12 12:15:02 · answer #6 · answered by Anonymous · 1⤊ 1⤋