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2 years ago

Part B

Mathematics

1 answer:

ziro4ka [17]2 years ago

3 0

The linear equation that has a slope of -7 and crosses the x-axis at (3, 0) is:

y = -7x + 21

<h3>How to find the linear equation?</h3>

A general linear equation is:

y = a*x +b

Where a is the slope and b is the y-intercept.

The slope must be equal to the limit found in part a, and you say that it is equal to -7, so the slope is -7. And for how is written the problem, I understand that it crosses the x-axis at x = 3.

Then we will have:

y = -7*x + b

Such that, when x = 3, y = 0, then:

0 = -7*3 + b

21 = b

Then the linear equation is y = -7x + 21

If you want to learn more about linear equations:

brainly.com/question/1884491

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Is the function shown linear or nonlinear? Explain.

marishachu [46]

Answer:

Step-by-step explanation:

The function is nonlinear because the slope is not constant.

5 0

3 years ago

There are 50 jelly beans. if 7 of them are red, and 23 are black, what percentage of the jelly beans are not reed

nadezda [96]

86% bc 50 is easy to work with, you know 43 arent red, and since 43x2=86 its 86, multiply by 2 bc 50 is half of 100

4 0

2 years ago

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

Honon spent 3 1/4 hours on homework while Sequoia spent 2 5/6 hours on homework. How much more time did Honon spend on homework

Rzqust [24]

Answer:

5/12 hours

Step-by-step explanation:

Honon = 3 1/4 hours

Sequoia = 2 5/6 hours

How much more time did Honon spend on homework than Sequoia

= number of hours Honon spent - number of hours Sequoia spent

= 3 1/4 hours - 2 5/6 hours

= 13/4 - 17/6

= (39-34) /12

= 5/12 hours

6 0

3 years ago

Si 15 obreros construyen un muro en 45 minutos, ¿en cuánto tiempo lo construirán 3 obreros con similar habilidad?

Alexxx [7]

Answer:

3 obreros tardaran 225 minutos en completar el trabajo.

Step-by-step explanation:

Sea R la velocidad con la que un obrero puede construir 1 muro.

Sabemos que 15 obreros construyen un muro en 45 minutos, entonces:

(15*R)*45min = 1 muro

Con esta ecuación podemos encontrar el valor R

R = (1 muro)/(15*45min)

Ahora queremos saber cuando tiempo tardan 3 obreros en construir un muro, entonces tenemos que resolver:

(3*R)*T = 1 muro

Reemplazando el valor de R obtenemos:

(3*(1 muro)/(15*45min))*T = 1 muro

T = (1 muro)*(15*45 min)/(3*1 muro)

T = 5*45min = 225 min

3 obreros tardaran 225 minutos en completar el trabajo.

7 0

3 years ago