Question

PLEASE HELP ASAP, WOULD APPRECIATE IF YOU SHOWED STEPS

Answer 1

1. a
2. b
3. b
4. b
5. c
6. a
7. b
8. (a)
   `7^(4)` (b)
   `((4)/(9)) ^(17)` (c)
   `4^(-4)` (d)
   `8^(-3)`

Explanation:

1. The moment he started: 3 bacteria

1st hour: 3*2 bacteria = 3*
`2^(1)` bacteria

2nd hour: 3*2*2 bacteria = 3*
`2^(2)` bacteria

3rd hour: 3*2*2*2 bacteria = 3*
`2^(3)` bacteria

etc.

Each hour the number of bacteria multiplies by 2 (the power of 2 is increased by 1), therefore,
`B=3*(2)^(n)`

2. Exponential function has the form of
`f(x)=ab^(x)`

if a is greater than 0 and b is greater than 1, then it's exponential growth.

The answer is:

`y=4^(x)` (4 > 1)

3. If smth increases by 2.5% each time, that means that at first you had 100% and then it became 100%+2.5% = 102.5%. So each time we multiply the value by 1.025 (100*1.025=102.5).

4. Linear equation form:
`y=ax+b`

Quadratic equation form:
`y=ax^(2) + bx + c`

Exponential equation form:
`y=ab^(x)`

In this case, a = 3.8, b = -5.5, c = -4.8

5. (a) square root

(b) linear

(c) exponential

(d) quadratic

6. Each year the car's worth is multiplied by 0.82 (82%). Therefore, after n years, the value will be: V = 27 500 *
`0.82^(n)`

7. Neither first, nor second differences are constant in case of an exponential relation, because we do not add but multiply.

Ex.
`y=2^(x)`

values: 1 2 4 8 16 32...

1st diff: 1 2 4 8 16...

2nd diff: 1 2 4 8...

8. (a)
`7^(11) / 7^(7) = 7^(11-7) = 7^(4)`

(b)
`((4)/(9))^(12) *((4)/(9))^(5) =((4)/(9) )^(12+5)=((4)/(9))^(17)`

(c)
`4^(11) / (4^(5))^(3) =4^(11) / 4^(5*3)=4^(11) /4^(15)=4^(11-15)=4^(-4)`

(d)
`8^(-6)*8^(3)=8^(-6+3)=8^(-3)`