QUESTION:
A battery was charged. When the charging began, it was 23 percent full. After 30 minutes of charging, the battery was 89 percent full. how fast did the battery charge and how long did it take?
Solution :
<u>Given</u>
- Initial amount of charge = 23%
- Amount after charge = 89%
<u>procedu</u><u>re</u>
Let the rate of charging be r .
r = (change in charge level) / (time interval) = (89% - 23%) / (30 min) = 2.2 % per minute
charge = (rate of charging)(time) + (initial charge)
100% = r*t + 23%
100% = (2.2 %/min)t + 23%
Solve for t and get
t = [(100 - 23) / 2.2]
Answer = 35 min
Answer:
3x−1
Step-by-step explanation:
1 Split the second term in 6{x}^{2}+x-16x
2
+x−1 into two terms.
\frac{6{x}^{2}+3x-2x-1}{4{x}^{2}-1}
4x
2
−1
6x
2
+3x−2x−1
2 Factor out common terms in the first two terms, then in the last two terms.
\frac{3x(2x+1)-(2x+1)}{4{x}^{2}-1}
4x
2
−1
3x(2x+1)−(2x+1)
3 Factor out the common term 2x+12x+1.
\frac{(2x+1)(3x-1)}{4{x}^{2}-1}
4x
2
−1
(2x+1)(3x−1)
4 Rewrite 4{x}^{2}-14x
2
−1 in the form {a}^{2}-{b}^{2}a
2
−b
2
, where a=2xa=2x and b=1b=1.
\frac{(2x+1)(3x-1)}{{(2x)}^{2}-{1}^{2}}
(2x)
2
−1
2
(2x+1)(3x−1)
5 Use Difference of Squares: {a}^{2}-{b}^{2}=(a+b)(a-b)a
2
−b
2
=(a+b)(a−b).
\frac{(2x+1)(3x-1)}{(2x+1)(2x-1)}
(2x+1)(2x−1)
(2x+1)(3x−1)
6 Cancel 2x+12x+1.
\frac{3x-1}{2x-1}
2x−1
3x−1
Answer:
sorry I don't know just ask to any one
Use a calculator app on your phone it’s not hard
-3(is less than)1 and 5(is less than or equal to) 25