Answer:
Your selection is appropriate
Step-by-step explanation:
A negative exponent in the numerator is equivalent to a positive exponent in the denominator, and vice versa.
... a⁻² = 1/a²
____
2⁴ multiplies the variable expression no matter which way it is written.
Answer:
b. (1, 3, -2)
Step-by-step explanation:
A graphing calculator or scientific calculator can solve this system of equations for you, or you can use any of the usual methods: elimination, substitution, matrix methods, Cramer's rule.
It can also work well to try the offered choices in the given equations. Sometimes, it can work best to choose an equation other than the first one for this. The last equation here seems a good one for eliminating bad answers:
a: -1 -5(1) +2(-4) = -14 ≠ -18
b: 1 -5(3) +2(-2) = -18 . . . . potential choice
c: 3 -5(8) +2(1) = -35 ≠ -18
d: 2 -5(-3) +2(0) = 17 ≠ -18
This shows choice B as the only viable option. Further checking can be done to make sure that solution works in the other equations:
2(1) +(3) -3(-2) = 11 . . . . choice B works in equation 1
-(1) +2(3) +4(-2) = -3 . . . choice B works in equation 2
Answer:
$40.20
Step-by-step explanation:
60 / 100 = 0.6
0.6 x 33 = 29.8
60 - 29.8 = 40.2
Thus he price after the discount would be $40.20
It's not compounded interest, you just get $3.20 every year.
$100 * 0.032 = $3.20
$3.20 * 3 years = $9.60
Answer:
10.5 hours.
Step-by-step explanation:
Please consider the complete question.
Working together, two pumps can drain a certain pool in 6 hours. If it takes the older pump 14 hours to drain the pool by itself, how long will it take the newer pump to drain the pool on its own?
Let t represent time taken by newer pump in hours to drain the pool on its own.
So part of pool drained by newer pump in one hour would be
.
We have been given that it takes the older pump 14 hours to drain the pool by itself, so part of pool drained by older pump in one hour would be
.
Part of pool drained by both pumps working together in one hour would be
.
Now, we will equate the sum of part of pool emptied by both pumps with
and solve for t as:








Therefore, it will take 10.5 hours for the newer pump to drain the pool on its own.