All categories

2 years ago

Can someone help me with this math homework please!

Mathematics

1 answer:

Delvig [45]2 years ago

3 0

Answer:

1. A -> k – 2 = 1/9(t – 18)

2. D -> Tim's with a y-intercept of $2,780

3. C -> y = x + 2

4. C -> y-8 = -0.2(x+10)

5. C -> $22.75

You might be interested in

Answer each question about each polygon.

oksian1 [2.3K]

the answer is a n b

Step-by-step explanation:

5 0

3 years ago

10.

timama [110]

Answer:

Step-by-step explanation:

3 0

3 years ago

Help please i will give 30 points

Nata [24]

Answer:

C. 7h + 50 > 99

Step-by-step explanation:

She earns <u>more</u> than $99, so you know that the sign will be > (more than).

h is the number of hours, so to find how much she earns for working h hours at $7/hour, you multiply h by 7 --- 7h

She also earns an additional $50, which will be +50.

Therefore, the equation is: 7h + 50 > 99

Hope this helps :D

5 0

3 years ago

What is the slope of the graph? please help ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

nignag [31]

Answer:

The slope of the graph is 3/5

Step-by-step explanation:

We know that slope is nothing but rise over run. In this equation, the spacing is 3 points different vertically and 5 horizontally. If we do Rise Over Run, we will get 3/5 as our answer.

5 0

3 years ago

Read 2 more answers

A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium d

nikitadnepr [17]

Answer:

a) k=2.08 1/hour

b) The exponential growth model can be written as:

$P(t)=Ce^{kt}$

c) 977,435,644 cells

d) 2.033 billions cells per hour.

e) 2.81 hours.

Step-by-step explanation:

We have a model of exponential growth.

We know that the population duplicates every 20 minutes (t=0.33).

The initial population is P(t=0)=58.

The exponential growth model can be written as:

$P(t)=Ce^{kt}$

For t=0, we have:

$P(0)=Ce^0=C=58$

If we use the duplication time, we have:

$P(t+0.33)=2P(t)\\\\58e^{k(t+0.33)}=2\cdot58e^{kt}\\\\e^{0.33k}=2\\\\0.33k=ln(2)\\\\k=ln(2)/0.33=2.08$

Then, we have the model as:

$P(t)=58e^{2.08t}$

The relative growth rate (RGR) is defined, if P is the population and t the time, as:

$RGR=\dfrac{1}{P}\dfrac{dP}{dt}=k$

In this case, the RGR is k=2.08 1/h.

After 8 hours, we will have:

$P(8)=58e^{2.08\cdot8}=58e^{16.64}=58\cdot 16,852,338= 977,435,644$

The rate of growth can be calculated as dP/dt and is:

$dP/dt=58[2.08\cdot e^{2.08t}]=120.64e^2.08t=2.08P(t)$

For t=8, the rate of growth is:

$dP/dt(8)=2.08P(8)=2.08\cdot 977,435,644 = 2,033,066,140$

(2.033 billions cells per hour).

We can calculate when the population will reach 20,000 cells as:

$P(t)=20,000\\\\58e^{2.08t}=20,000\\\\e^{2.08t}=20,000/58\approx344.827\\\\2.08t=ln(344.827)\approx5.843\\\\t=5.843/2.08\approx2.81$

3 0

3 years ago