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3 years ago

True or false? Postulates are statements that require proof

Mathematics

2 answers:

viktelen [127]3 years ago

7 0

False. Postulates are statements that are accepted without question or justification.

Kryger [21]3 years ago

6 0

Answer:

The statement is FALSE.

Step-by-step explanation:

Postulates or also called axioms, are taken to be true without proof. These are statements that are considered to be self-evident.

Few examples of postulates are -

1. A straight line segment may be drawn from any given point to any other.

2. A straight line may be extended to any finite length.

3. All right angles are congruent.

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find the area of the circle (test grade due in 30 min) use 3.14. round your answer to the nearest hundredth.

DENIUS [597]

Answer:

πr²

992.25

Step-by-step explanation:

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3 years ago

Can someone help me with the question?

vladimir2022 [97]

Corndogs broo likeee everyone knows this

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3 years ago

Jim decides to start a small nonprofit business of renting out motor scooters to help out his area residents. He puts in his own

Deffense [45]

First, find the expected number of scooters rented per month:

As the data is symmetrical, E(X) (the expected value) is the middle value. So, on average, 2.5 scooters should be taken per month.

His total costs were 5 * 3000 = $15,000

So, to break even, he needs to make $15,000.

He will be selling for 5 years, or 60 months.

As a result, he needs to make 15000/60 = $250/month

As he is selling 2.5 scooters on average, he needs to rent each for:

$250/2.5 = <u>$100/month</u>

6 0

4 years ago

Read 2 more answers

Find a second solution y2(x) of<br> x^2y"-3xy'+5y=0; y1=x^2cos(lnx)

rosijanka [135]

We can try reduction order and look for a solution $y_2(x)=y_1(x)v(x)$ . Then

$y_2=y_1v\implies{y_2}'=y_1v'+{y_1}'v\implies{y_2}''=y_1v''+2{y_1}'v+{y_1}''v$

Substituting these into the ODE gives

$x^2(y_1v''+2{y_1}'v+{y_1}''v)-3x(y_1v'+{y_1}'v)+5y_1v=0$

$x^2y_1v''+(2x^2{y_1}'-3xy_1)v'+(x^2{y_1}''-3x{y_1}'+5y_1)v=0$

$x^4\cos(\ln x)v''+x^3(\cos(\ln x)-2\sin(\ln x))v'=0$

which leaves us with an ODE linear in $w(x)=v'(x)$ :

$x^4\cos(\ln x)w'+x^3(\cos(\ln x)-2\sin(\ln x))w=0$

This ODE is separable; divide both sides by the coefficient of $w'(x)$ and separate the variables to get

$w'+\dfrac{\cos(\ln x)-2\sin(\ln x)}{x\cos(\ln x)}w=0$

$\dfrac{w'}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}$

$\dfrac{\mathrm dw}w=\dfrac{2\sin(\ln x)-\cos(\ln x)}{x\cos(\ln x)}\,\mathrm dx$

Integrate both sides; on the right, substitute $u=\ln x$ so that $\mathrm du=\dfrac{\mathrm dx}x$ .

$\ln|w|=\displaystyle\int\frac{2\sin u-\cos u}{\cos u}\,\mathrm du=\int(2\tan u-1)\,\mathrm du$

Now solve for $w(u)$ ,

$\ln|w|=-2\ln(\cos u)-u+C$

$w=e^{-2\ln(\cos u)-u+C}$

$w=Ce^{-u}\sec^2u$

then for $w(x)$ ,

$w=Ce^{-\ln x}\sec^2(-\ln x)$

$w=C\dfrac{\sec^2(\ln x)}x$

Solve for $v(x)$ by integrating both sides.

$v=\displaystyle C_1\int\frac{\sec^2(\ln x)}x\,\mathrm dx$

Substitute $u=\ln x$ again and solve for $v(u)$ :

$v=\displaystyle C_1\int\sec^2u\,\mathrm du$

$v=C_1\tan u+C_2$

then for $v(x)$ ,

$v=C_1\tan(\ln x)+C_2$

So the second solution would be

$y_2=x^2\cos(\ln x)(C_1\tan(\ln x)+C_2)$

$y_2=C_1x^2\sin(\ln x)+C_2x^2\cos(\ln x)$

$y_1(x)$ already accounts for the second term of the solution above, so we end up with

$\boxed{y_2=x^2\sin(\ln x)}$

as the second independent solution.

6 0

4 years ago

A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation

melamori03 [73]

Answer:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The point estimate of the population mean is $\hat \mu = \bar X In order to calculate the critical value [tex]t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=64-1=63$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=1400$ represent the sample mean

$\mu$ population mean (variable of interest)

s=240 represent the sample standard deviation

n=64 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The point estimate of the population mean is $\hat \mu = \bar X In order to calculate the critical value [tex]t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=64-1=63$

5 0

4 years ago