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Kazeer [188]
4 years ago
13

What are the values of v and b? (geometry question)

Mathematics
1 answer:
Mnenie [13.5K]4 years ago
3 0

Answer:

b=41

Step-by-step explanation:

180=19v-41+(180-(20v+39))+17v

v=5

180=19v-41+b+17v

180=19(5)-41+b+17(5)

b=41

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The population of a certain town was 10,000 in 1990. The rate of change of the population, measured in people per year, is model
Katena32 [7]

The question is incomplete. The complete question is :

The population of a certain town was 10,000 in 1990. The rate of change of a population, measured in hundreds of people per year, is modeled by P prime of t equals two-hundred times e to the 0.02t power, where t is measured in years since 1990. Discuss the meaning of the integral from zero to twenty of P prime of t, d t. Calculate the change in population between 1995 and 2000. Do we have enough information to calculate the population in 2020? If so, what is the population in 2020?

Solution :

According to the question,

The rate of change of population is given as :

$\frac{dP(t)}{dt}=200e^{0.02t}$  in 1990.

Now integrating,

$\int_0^{20}\frac{dP(t)}{dt}dt=\int_0^{20}200e^{0.02t} \ dt$

                    $=\frac{200}{0.02}\left[e^{0.02(20)}-1\right]$

                   $=10,000[e^{0.4}-1]$

                    $=10,000[0.49]$

                    =4900

$\frac{dP(t)}{dt}=200e^{0.02t}$

$\int1.dP(t)=200e^{0.02t}dt$

$P=\frac{200}{0.02}e^{0.02t}$

$P=10,000e^{0.02t}$

$P=P_0e^{kt}$

This is initial population.

k is change in population.

So in 1995,

$P=P_0e^{kt}$

   $=10,000e^{0.02(5)}$

   $=11051$

In 2000,

$P=10,000e^{0.02(10)}$

   =12,214

Therefore, the change in the population between 1995 and 2000 = 1,163.

8 0
3 years ago
What is the slope of the line that passes through the point (9,4) and (3,9)?​
Fantom [35]

Step-by-step explanation:

m is for slope of the line that passes through

m =  \frac{y_2 - y_1}{x_2 - x_1}

m =  \frac{9 - 4}{3 - 9}

m =  \frac{5}{ - 6}  =  -  \frac{5}{6}

8 0
2 years ago
Given f (x) = 2x^2 -3x + 5, find f (-4)
bogdanovich [222]
-1=x is the answer I’m pretty sure
4 0
3 years ago
Read 2 more answers
Year: %Households:
harkovskaia [24]

Based on the table showing the percentage of households playing games over the net, the average rate of change from 1999 to 2003 is 3.9% per year.

<h3>What is average rate of change?</h3>

This can be found as:

= (27.9 - 12.3) / 4 years

= 3.9% per year

In 2000, the instantaneous rate of change would be:

= (Rate in 2001 - Rate in 1999) / difference in years

= (24.4 - 12.3) / (2001 - 1999)

= 6.05%

Find out more on the average rate of change at brainly.com/question/2263931.

#SPJ1

6 0
2 years ago
Round 56,477,812 to the nearest hundred thousand,
GrogVix [38]

Answer:

56,500,000.

Step-by-step explanation:

56,477,812 rounded to the nearest hundred thousand:

The 4 is in the hundred thousands place, so we'll look at the next digit to the right of that, which is the 7:

56,<u>4</u>77,812

Since 7 is more than 5, we'll have to go up a number, which will be the 4. Afterwards, we'll have to replace all the digits after the 4 with zeros.

56,500,000.

8 0
3 years ago
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