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PLEASE HELP ASAP, WOULD APPRECIATE IF YOU SHOWED STEPS

PLEASE HELP ASAP, WOULD APPRECIATE IF YOU SHOWED STEPS-example-1
User Arjun Raj
by
8.2k points

1 Answer

6 votes

Answer:

  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)

User Martin Seubert
by
8.9k points

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