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3 years ago

If f(x)=4x+7 and g(x)=x^3, what is (g°f)(-1)

Mathematics

1 answer:

Annette [7]3 years ago

7 0

Answer:

The answer is (g°f)(-1) = 27

Step-by-step explanation:

* (g°f)(1) ⇒ means make the domain of g is the range of f

- At first find the value of f(-1)

- Take the answer and substitute it as the value of x for g

- So the range of f is the domain of g

∵ f(x) = 4x + 7

∵ The domain of f = -1 ⇒ x is the domain

∴ f(-1) = 4(-1) + 7 = -4 + 7 = 3 ⇒ f(-1) is the range

∵ g(x) = x³

∵ x = f(-1) = 3 ⇒ the domain of g

∴ g(3) = 3³ = 27 ⇒ the range of g

* (g°f)(-1) = 27

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The standard deviation of a sample taken from population A is 17.6 for a sample of 25.

lawyer [7]

Answer:

The standard deviation of the sample mean differences is _5.23_

Step-by-step explanation:

We have a sample of a population A and a sample of a population B.

For the sample of population A, the standard deviation $\sigma_A$ is

$\sigma_A = 17.6$

The sample size $n_A$ is:

$n_A = 25$ .

For the sample of population B, the standard deviation $\sigma_B$ is

$\sigma_B = 21.2$

The sample size $n_B$ is:

$n_B = 30$ .

Then the standard deviation for the difference of means has the following form:

$\sigma=\sqrt{\frac{\sigma_A^2}{n_A}+\frac{\sigma_B^2}{n_B}}$

Finally

$\sigma=\sqrt{\frac{17.6^2}{25}+\frac{21.2^2}{30}}\\\\\sigma= 5.23$

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Answer:

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Step-by-step explanation:

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Answer:

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mariarad [96]

The first equation is linear:

$x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x$

Divide through by $x^2$ to get

$\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x$

and notice that the left hand side can be consolidated as a derivative of a product. After doing so, you can integrate both sides and solve for $y$ .

$\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x$
$\implies\dfrac1xy=\displaystyle\int\sin x\,\mathrm dx=-\cos x+C$
$\implies y=-x\cos x+Cx$

- - -

The second equation is also linear:

$x^2y'+x(x+2)y=e^x$

Multiply both sides by $e^x$ to get

$x^2e^xy'+x(x+2)e^xy=e^{2x}$

and recall that $(x^2e^x)'=2xe^x+x^2e^x=x(x+2)e^x$ , so we can write

$(x^2e^xy)'=e^{2x}$
$\implies x^2e^xy=\displaystyle\int e^{2x}\,\mathrm dx=\frac12e^{2x}+C$
$\implies y=\dfrac{e^x}{2x^2}+\dfrac C{x^2e^x}$

- - -

Yet another linear ODE:

$\cos x\dfrac{\mathrm dy}{\mathrm dx}+\sin x\,y=1$

Divide through by $\cos^2x$ , giving

$\dfrac1{\cos x}\dfrac{\mathrm dy}{\mathrm dx}+\dfrac{\sin x}{\cos^2x}y=\dfrac1{\cos^2x}$
$\sec x\dfrac{\mathrm dy}{\mathrm dx}+\sec x\tan x\,y=\sec^2x$
$\dfrac{\mathrm d}{\mathrm dx}[\sec x\,y]=\sec^2x$
$\implies\sec x\,y=\displaystyle\int\sec^2x\,\mathrm dx=\tan x+C$
$\implies y=\cos x\tan x+C\cos x$
$y=\sin x+C\cos x$

- - -

In case the steps where we multiply or divide through by a certain factor weren't clear enough, those steps follow from the procedure for finding an integrating factor. We start with the linear equation

$a(x)y'(x)+b(x)y(x)=c(x)$

then rewrite it as

$y'(x)=\dfrac{b(x)}{a(x)}y(x)=\dfrac{c(x)}{a(x)}\iff y'(x)+P(x)y(x)=Q(x)$

The integrating factor is a function $\mu(x)$ such that

$\mu(x)y'(x)+\mu(x)P(x)y(x)=(\mu(x)y(x))'$

which requires that

$\mu(x)P(x)=\mu'(x)$

This is a separable ODE, so solving for $\mu$ we have

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and so on.

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