Answer:
The answer is the last one (32x^7y^15)
You can bring x to the second power (x^2) because (x) is basically x^1. This is a basic exponent rule. (x^m)^n = x^m times n.
Then you can apply this rule to (2xy^3)^5. First you bring two to the fifth power and get 32. Then you bring x^5 according to the rule. Then you bring y^15, also because of the rule.
Now you have:
x^2 times 32x^5y^15
Now you just multiply the like terms together (x^2 and x^5)
When you multiply two exponents with the same base, you add the exponents together: a^n times a^m = a^n+m.
So you end up with 32x^7y^15
Use a let statement
first
let x and x + 2 be the number so you write it like this
<u>let x = the first consecutive integer
</u><u>let x + 2 = the second consecutive integer
</u>
second
x(x+2)=323
x^2 + 2x = 323
-323 -323
x^2 + 2x -323 = 0
third
try to factor -323 so it is 19 and -17
(x + 19) (x - 17) = 0
x = 19
x = -17
hope this help
Answer:
25/16
Step-by-step explanation:
(4/5)^−2
=
(4/5)^−2
=
(5/4)^2
=
(5/4)*(5/4)
=
5*5/4*4
=
5^2/4^2
=
25/16
(Decimal: 1.5625)
