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3 years ago

9

A single bacterium lands in your mouth and starts growing by a factor of 4 every hour. After how many hours will the number of b

acteria exceed 1,000?

1 answer:

Ierofanga [76]3 years ago

8 0

Answer:

After 5 hours, the number of bacteria will be 1024

Step-by-step explanation:

1 times 4 = 4

4 times 4 = 16

16 times 4 = 64

64 times 4 = 256

256 times 4 = 1024

It will take around 5 hours to exceed 1000

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Students are given 3 minutes to complete each multiple-choice question on a test and 8 minutes for each free-response question.

Alja [10]

The value that could replace variable x in the table is (15 - m)8

<h3 /><h3>How to the value of x in an expression ?</h3>

The students are given 3 minutes to complete each multiple choice question on a test and 8 minutes for each free response question.

From the table, the time to answer the whole multiple choice question is as follows:

total time = 3m

From the table, the total time taken to answer the free response question is as follows:

total time = (15 - m)8

learn more on variable here: brainly.com/question/12798965

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5 0

2 years ago

Is 5 in the solution set x +3≥8

forsale [732]

Answer: Yes

Step-by-step explanation:

x>=8-3

x>=5

6 0

3 years ago

Friday, a store sold 10 blue shirts and 20 white shirts. what percentage of the shirts sold were blue? give answer to the ones p

jasenka [17]

Answer:

10+20=30

10 is what percent of 30?

10 is 1/3 of 30

Step-by-step explanation:

10 is 33.33% of 30

so 33% of the shirts sold were blue

6 0

3 years ago

3 Write a numerical expression to model the words. Subtract 1 from the product of 4 and 5.

umka21 [38]

Answer:

Hello!

After reading the question you have provided I have come up with the correct numerical expression:

4x5-1

Step-by-step explanation:

To come up with this solution you need to keep in mind some of the terminoloy being used.

The word "subtract" comes from the action of subtraction

The word "product" comes from the action of multiplication

Thus, using those terminologies correctly, you can then deduce that when the question says "the product of 4 and 5" means "multiplying 4 and 5 together".

So you get the first part being 4x5

Then, you add in the last part of "subract 1" from the "product of 4 and 5":

4x5-1

<em>Remember to keep in mind the rule of "PEMDAS"</em>

You always need to keep the multiplication portion of the equation in front of any subtraction, or addition in any given equation.

7 0

3 years ago

Read 2 more answers

21. Who is closer to Cameron? Explain.

pickupchik [31]

Problem 21

<h3>Answer: Jamie is closer</h3>

-----------------------

Explanation:

A = Arthur's location = (20,35)
J = Jamie's location = (45,20)
C = Cameron's location = (65,40)

To find out who's closer to Cameron, we need to compute the segment lengths AC and JC. Then we pick the smaller of the two lengths.

We use the distance formula to find each length

Let's find the length of AC.

$A = (x_1,y_1) = (20,35)\\\\C = (x_2,y_2) = (65,40)\\\\d = \text{Distance from A to C} = \text{length of segment AC}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(20-65)^2 + (35-40)^2}\\\\d = \sqrt{(-45)^2 + (-5)^2}\\\\d = \sqrt{2025 + 25}\\\\d = \sqrt{2050}\\\\d \approx 45.2769257\\\\$

The distance from Arthur to Cameron is roughly 45.2769257 units.

Let's repeat this process to find the length of segment JC

$J = (x_1,y_1) = (45,20)\\\\C = (x_2,y_2) = (65,40)\\\\d = \text{Distance from J to C} = \text{length of segment JC}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(45-65)^2 + (20-40)^2}\\\\d = \sqrt{(-20)^2 + (-20)^2}\\\\d = \sqrt{400 + 400}\\\\d = \sqrt{800}\\\\d \approx 28.2842712\\\\$

Going from Jamie to Cameron is roughly 28.2842712 units

We see that segment JC is shorter than AC. Therefore, Jamie is closer to Cameron.

=================================================

Problem 22

<h3>Answer: Arthur is closest to the ball</h3>

-----------------------

Explanation:

We have these key locations:

A = Arthur's location = (20,35)
J = Jamie's location = (45,20)
C = Cameron's location = (65,40)
B = location of the ball = (35,60)

We'll do the same thing as we did in the previous problem. This time we need to compute the following lengths:

AB
JB
CB

These segments represent the distances from a given player to the ball. Like before, the goal is to pick the smallest of these segments to find out who is the closest to the ball.

The steps are lengthy and more or less the same compared to the previous problem (just with different numbers of course). I'll show the steps on how to get the length of segment AB. I'll skip the other set of steps because there's only so much room allowed.

$A = (x_1,y_1) = (20,35)\\\\B = (x_2,y_2) = (35,60)\\\\d = \text{Distance from A to B} = \text{length of segment AB}\\\\d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}\\\\d = \sqrt{(20-35)^2 + (35-60)^2}\\\\d = \sqrt{(-15)^2 + (-25)^2}\\\\d = \sqrt{225 + 625}\\\\d = \sqrt{850}\\\\d \approx 29.1547595\\\\$

Segment AB is roughly 29.1547595 units.

If you repeated these steps, then you should get these other two approximate segment lengths:

JB = 41.2310563

CB = 36.0555128

-------------

So in summary, we have these approximate segment lengths

AB = 29.1547595
JB = 41.2310563
CB = 36.0555128

Segment AB is the smallest of the trio, which therefore means Arthur is closest to the ball.

3 0

3 years ago