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3 years ago

4.9-16=. A:25 B:7 C:-25 D:-7

Mathematics

1 answer:

shtirl [24]3 years ago

8 0

Answer and Step-by-step explanation:

It says your question is : $9-16=?$ And the answer for that would be $-10$ . But that's not there in the options I think your question was supposed to be meant like this : $16-9=?$ cause then it would be $7$ tho it's there in the options, the sign is wrong it's not $-7$ it's $+7$ or just $7$ .

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Find the volume of the figure:<br>7ft, 2ft, 1ft, 6ft, 7ft, 6ft, 2ft, 1ft,

andreev551 [17]

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3 years ago

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OlgaM077 [116]

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3 years ago

given y>0 and dy/dx=(3x^2+4x)/yif the point (1,√10) is on the graph relating x and y, then what is y when x=0

jonny [76]

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{3x^2+4x}y\iff y\,\mathrm dy=(3x^2+4x)\,\mathrm dx$
$\implies\displaystyle\int y\,\mathrm dy=\int(3x^2+4x)\,\mathrm dx$
$\implies \dfrac12y^2=x^3+2x^2+C$

When $x=1$ you have $y=\sqrt{10}$ , so

$\dfrac12(\sqrt{10})^2=1^3+2(1)^2+C\implies 5=1+2+C\implies C=2$

and so the particular solution to the ODE is

$\dfrac12y^2=x^3+2x^2+2$

Then when $x=0$ , you get

$\dfrac12y^2=0^3+2(0)^2+2=2$
$\implies y^2=4$
$\implies y=2$

where we omitted the negative root because it's given that $y>0$ .

4 0

3 years ago

Find the indefinite integral. (Note: Solve by the simplest method—not all require integration by parts. Use C for the constant o

umka2103 [35]

Answer:

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2 ln⁡(1+x^2 )+C

Step-by-step explanation:

∫▒〖1st .2nd dx=1st∫▒〖2nd dx〗-∫▒〖(derivative of 1st) dx∫▒〖2nd dx〗〗〗

Let 1st=arctan⁡(x)

And 2nd=1

∫▒〖arctan⁡(x).1 dx=arctan⁡(x) ∫▒〖1 dx〗-∫▒〖(derivative of arctan(x))dx∫▒〖1 dx〗〗〗

As we know that

derivative of arctan(x)=1/(1+x^2 )

∫▒〖1 dx〗=x

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-∫▒〖(1/(1+x^2 ))dx.x〗…………Eq1

Let’s solve ∫▒(1/(1+x^2 ))dx by substitution now

Let 1+x^2=u

du=2xdx

Multiply and divide ∫▒〖(1/(1+x^2 ))dx.x〗 by 2 we get

1/2 ∫▒〖(2/(1+x^2 ))dx.x〗=1/2 ∫▒(2xdx/u)

1/2 ∫▒(2xdx/u) =1/2 ∫▒(du/u)

1/2 ∫▒(2xdx/u) =1/2 ln⁡(u)+C

1/2 ∫▒(2xdx/u) =1/2 ln⁡(1+x^2 )+C

Putting values in Eq1 we get

∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2 ln⁡(1+x^2 )+C (required soultion)

3 0

3 years ago

Read 2 more answers

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

Leni [432]

Answer: $\dfrac{2x^2-1}{x(x^2-1)}$

Step-by-step explanation:

The given function : $y=\ln(x(x^2 - 1)^{\frac{1}{2}})$

$\Rightarrow\ y=\ln x+\ln (x^2-1)^{\frac{1}{2}}$ [ $\because \ln(ab)=\ln a +\ln b$ ]

$\Rightarrow y=\ln x+\dfrac{1}{2}\ln (x^2-1)}$ [ $\because \ln(a)^n=n\ln a$ ]

Now , Differentiate both sides with respect to x , we will get

$\dfrac{dy}{dx}=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})\dfrac{d}{dx}(x^2-1)$ (By Chain rule)

[Note : $\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}$ ]

$\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x-0)$

[ $\because \dfrac{d}{dx}(x^n)=nx^{n-1}$ ]

$=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x) = \dfrac{1}{x}+\dfrac{x}{x^2-1}\\\\\\=\dfrac{(x^2-1)+(x^2)}{x(x^2-1)}\\\\\\=\dfrac{2x^2-1}{x(x^2-1)}$

Hence, the derivative of the given function is $\dfrac{2x^2-1}{x(x^2-1)}$ .

8 0

4 years ago