All categories

3 years ago

In ΔRST, s = 7.3 inches, t = 3.1 inches and ∠R=138°. Find the area of ΔRST, to the nearest 10th of a square inch.

Mathematics

1 answer:

Oksana_A [137]3 years ago

8 0

Answer:

5.1 inches

Step-by-step explanation:

Five triangles make up the pentagon.

The area of each is aside

The area of the pentagon is 5

Substitute values and calculate

You might be interested in

Rita earns $10 per hour. She puts 5% of her earnings in savings. Write inequality to find how many hours Rita must work to save

Rina8888 [55]

Answer:

Rita must work 50 days to save at least $25.

Step-by-step explanation:

The amount Rita earns per hour = $10

The saving percentage = 5%

So, the savings of Rita per hour = 5% of $10

⇒5% of $10 = $\frac{5}{100} \times 10 = 0.5$

or Rita saves $0.5 per hour.

The amount she wants to save at minimum = $25

So, let she works at the minimum of k Days.

⇒ $\textrm{The number of hours she has to work} = \frac{\textrm{Amount she needs saving}}{\textrm{Amount saved in 1 hour}}$

or, $k = \frac{25}{0.5} = 50$

The number of hours she has to work = 50 days

Hence, Rita must work 50 days to save at least $25.

3 0

3 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

In △ABC, point P∈

lorasvet [3.4K]

<u><em> $\beta \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \lim_{n \to \infty} a_n \lim_{n \to \infty} a_n x_{123} \left \{ {{y=2} \atop {x=2}} \right. \left \{ {{y=2} \atop {x=2}} \right. \alpha \alpha \sqrt{x} \sqrt{x}$ </em></u>