Answer:
x=34
Step-by-step explanation:
6 - ( x-7) ^ 1/3 = 3
Subtract 6 from each side
6-6 - ( x-7) ^ 1/3 = 3-6
- ( x-7) ^ 1/3 = -3
Divide each side by a negative
( x-7) ^ 1/3 = 3
Cube each side
( x-7) ^ 1/3 ^3 = (3)^3
x-7 = 27
Add 7 to each side
x-7+7 = 27+7
x = 34
Check
6 - ( 34-7) ^ 1/3 = 3
6 - (27^1/3 = 3
6 -3 =3
3=3
Good solution
Answer: 7x^2+21x+14
Step-by-step explanation:
(7x+7)(x+2)
Multiply each term in the first parentheses by each term in the second parentheses (FOIL)
7x×x+7x×2+7x+7×2
↘ ↙
7x×x calculate product
7x^2+7x×2+7x+7×2
↘ ↙
7x×2 calculate product
7x^2+14x+7x+7×2
↘↙
7×2 multiply numbers
7x^2+14x+7x+14
↘ ↙
21x collect like terms
7x^2+21x+14 is your end result.
Not exactly sure if I’m right but
1. $56 < $60
2. $60 - $36 = $24/4 hours = 6.
36+24 = 60, we want to use LESS than 60 so we use see how much it’ll take to spend 5 hours for parking.
5 x $4 = $20.
$36 + $20 = $56
I’m not exactly sure though...
Answer:
- P(t) = 100·2.3^t
- 529 after 2 hours
- 441 per hour, rate of growth at 2 hours
- 5.5 hours to reach 10,000
Step-by-step explanation:
It often works well to write an exponential expression as ...
value = (initial value)×(growth factor)^(t/(growth period))
(a) Here, the growth factor for the bacteria is given as 230/100 = 2.3 in a period of 1 hour. The initial number is 100, so we can write the pupulation function as ...
P(t) = 100·2.3^t
__
(b) P(2) = 100·2.3^2 = 529 . . . number after 2 hours
__
(c) P'(t) = ln(2.3)P(t) ≈ 83.2909·2.3^t
P'(2) = 83.2909·2.3^2 ≈ 441 . . . bacteria per hour
__
(d) We want to find t such that ...
P(t) = 10000
100·2.3^t = 10000 . . . substitute for P(t)
2.3^t = 100 . . . . . . . . divide by 100
t·log(2.3) = log(100)
t = 2/log(2.3) ≈ 5.5 . . . hours until the population reaches 10,000
Answer:
b. x = 80, y = 100, z = 63
Step-by-step explanation:
