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3 years ago

What is the value of s?

Mathematics

1 answer:

KonstantinChe [14]3 years ago

6 0

Answer:

35°

Step-by-step explanation:

Inscribed angle s is half the measure of the arc it subtends. That arc is the supplement to the 110° arc shown. The arc is 70°, so angle s is 35°.

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Pumping stations deliver oil at the rate modeled by the function D, given by d of t equals the quotient of 5 times t and the qua

goblinko [34]

<h2>Hello!</h2>

The answer is: There is a total of 5.797 gallons pumped during the given period.

To solve this equation, we need to integrate the function at the given period (from t=0 to t=4)

The given function is:

$D(t)=\frac{5t}{1+3t}$

So, the integral will be:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dx$

So, integrating we have:

$\int\limits^4_0 {\frac{5t}{1+3t}} \ dt=5\int\limits^4_0 {\frac{t}{1+3t}} \ dx$

Performing a change of variable, we have:

$1+t=u\\du=1+3t=3dt\\x=\frac{u-1}{3}$

Then, substituting, we have:

$\frac{5}{3}*\frac{1}{3}\int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du\\\\\frac{5}{9} \int\limits^4_0 {\frac{u-1}{u}} \ du=\frac{5}{9} \int\limits^4_0 {\frac{u}{u} -\frac{1}{u } \ du$

$\frac{5}{9} \int\limits^4_0 {(\frac{u}{u} -\frac{1}{u } )\ du=\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u } )$

$\frac{5}{9} \int\limits^4_0 {(1 -\frac{1}{u })\ du=\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du$

$\frac{5}{9} \int\limits^4_0 {(1 )\ du- \frac{5}{9} \int\limits^4_0 {(\frac{1}{u })\ du=\frac{5}{9} (u-lnu)/[0,4]$

Reverting the change of variable, we have:

$\frac{5}{9} (u-lnu)/[0,4]=\frac{5}{9}((1+3t)-ln(1+3t))/[0,4]$

Then, evaluating we have:

$\frac{5}{9}((1+3t)-ln(1+3t))[0,4]=(\frac{5}{9}((1+3(4)-ln(1+3(4)))-(\frac{5}{9}((1+3(0)-ln(1+3(0)))=\frac{5}{9}(10.435)-\frac{5}{9}(1)=5.797$

So, there is a total of 5.797 gallons pumped during the given period.

Have a nice day!

4 0

3 years ago

Writing Equations of Parallel Lines

Nookie1986 [14]

Answer:

The slope of the parallel line to the given line is $-\frac{1}{4}$

The equation of the parallel line to the given line and passes through the given point is y + 4 = $-\frac{1}{4}$ (x + 2)

The y-intercept of the parallel line to the given line and passes through the given point is $-\frac{9}{2}$

Step-by-step explanation:

The rule of the slope of the line that passes through points (x1, y1) and (x2, y2) is m = $\frac{y2-y1}{x2-x1}$

The point-slope form of the linear equation is y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line

The slope-intercept form of the linear equation is y = m x + b, where m is the slope and b is the y-intercept

Parallel lines have the same slopes and different y-intercepts

In the given figure

∵ The given line passes through points (2, 6) and (-6, 8)

∴ x1 = 2 and y1 = 6

∴ x2 = -6 and y2 = 8

→ Substitute them in the rule of the slope above to find it

∵ m = $\frac{8-6}{-6-2}$ = $\frac{2}{-8}$ = $-\frac{1}{4}$

∴ The slope of the given line is $-\frac{1}{4}$

∵ Parallel lines have the same slopes

∴ The slope of the parallel line to the given line is $-\frac{1}{4}$

∵ The parallel line passes through the point (-2, -4)

∴ x1 = -2 and y1 = -4

∵ m = $-\frac{1}{4}$

→ Substitute them in the point-slope form above

∵ y - (-4) = $-\frac{1}{4}$ (x - (-2))

∴ y + 4 = $-\frac{1}{4}$ (x + 2)

∴ The equation of the parallel line to the given line and passes through

the given point is y + 4 = $-\frac{1}{4}$ (x + 2)

∵ m = $-\frac{1}{4}$

→ Substitute it in the slope-intercept form above

∴ y = $-\frac{1}{4}$ x + b

→ To find b substitute x by -2 and y by -4 (coordinates the given point)

∵ -4 = $-\frac{1}{4}$ (-2) + b

∴ -4 = $\frac{1}{2}$ + b

→ Subtract $\frac{1}{2}$ from both sides

∴ $-\frac{9}{2}$ = b

∵ b is the y-intercept

∴ The y-intercept of the parallel line to the given line and passes

through the given point is $-\frac{9}{2}$

8 0

3 years ago

An accounting professor wants to know the average GPA of the students enrolled in her class. She looks up information on Blackbo

olya-2409 [2.1K]

Answer:

a) True

b) Parameter

Step-by-step explanation:

We are given the following in the question:

An accounting professor wants to know the average GPA of the students enrolled in her class.

Population:

It is defined as the collection of all variables of interest.
A sample of individuals of interest is drawn from a population.

For the given case

Individuals of interest:

Students enrolled in accounting class

Population of interest:

Students enrolled in accounting class

Characteristic of interest:

Average GPA of the students enrolled in her class.

a) The population is all students enrolled in the accounting class.

The given statement is true as it contains all the observation of all the individuals of interest.

b) The computed average GPA of all the students enrolled in the class is 3.29

Statistic is a descriptive measure that describes a sample where as a parameter is a measure that describes the population.

Since 3.29 is the average of all the students enrolled in her class that is the average GPA of population.

Thus, 3.29 is a parameter.

6 0

4 years ago

Consider the growth of a population p(t). It starts out with p(0)=A. Suppose the growth is unchecked, and hence p′=kp for some c

gayaneshka [121]

Answer:

$p(t) = Ae^{\frac{kt^2}{2}}$

Step-by-step explanation:

We are given the following information in the question:

The growth of population is given by the function p(t).

p(0)=A where A is a constant.

$p' = kp$

where k is a constant.

Solving the given differential equation, we have,

$p' = kp\\\\\displaystyle\frac{dp}{dt} = kp\\\\\frac{dp}{p} = kt~ dt\\\\\text{Integrating both sides}\\\\\int \frac{dp}{p} = \int kt~ dt\\\\\log p = \frac{kt^2}{2} + C\\\\\text{where C is the integration constant}\\\\\text{Putting t = 0}\\\\\log p_0 = C\\\\\log p = \frac{kt^2}{2} + \log p_0\\\\\log p - \log A = \frac{kt^2}{2}\\\\\log \frac{p}{A} = \frac{kt^2}{2}\\\\p(t) = Ae^{\frac{kt^2}{2}}$

5 0

3 years ago

Can someone help me on my algebra 1 ?

kramer

Sure comment down the question.

4 0

3 years ago