There are no 'real' numbers that can do the job.

The numbers that CAN are

8.5 + 1.66i
and
8.5 - 1.66i.

These are called 'complex' numbers.
The ' i ' is √-1 .

6 0

3 years ago

Read 2 more answers

I need help to solve this and also Hearn will the bacteria be over 100,000

lapo4ka [179]

Hello! Let's look at the two parts of this question.

Complete the table:

In this case, you just substitute the value of "hour" into the equation, for the value of t. For example:

P(0) = 120 $(2)^{0}$

P(0) = 120 (1)

P(0) = 120

Therefore, the number of bacteria for hour 0 is 120.

You can do this for the next ones. Hour 1 = 240, hour 2 = 480, and so on. (In this case, you can keep multiplying by 2)

Estimate when there will be more than 100,000 bacteria:

Set the final value of P(t) = 100,000, then solve.

100,000 = 120 (2 $)^{t}$

833.33 = (2 $)^{t}$

t = $log_{2}833.33$

t = 9.702744108

So your answer would be around 9.7 years, or, around 10 years.

Hope this helps!

3 0

3 years ago

Arrange the steps in the correct order to perform this subtraction problem 1.1 ( 10 3 ) − 4.9 ( 10 2 ) = 610

Norma-Jean [14]

We are given : Expression 1.1 ( 10^3 ) − 4.9 ( 10^2 ) = 610.

Solution: In order to solve above given expression, we need to apply order of operations.

We can apply order of operations according to PEMDAS rule.

PEMDAS stands for Parentheses, Exponents, Multiplication/Division, Addition/Subtraction.

In the given problem, we need to simplify Parentheses first.

We have 10^3 and 10^2 in Parentheses.

10^3 = 10*10*10 = 1000.

10^2 = 10*10 = 100.

Substituting those values in original expression.

1.1 ( 1000 ) − 4.9 ( 100 ).

Now, we need to apply multiplication operation in 1.1*1000 and 4.9*100.

We get 1.1*1000 = 1100 and 4.9*100 = 490.

Substituting those values in above result we got.

1100-490.

Now, we can perform subtraction operation.

On subtracting 1100-490, we get 610.

And that is equal to 610.

7 0

3 years ago