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3 years ago

Find the geometric mean of 2 and 8

Mathematics

2 answers:

-BARSIC- [3]3 years ago

3 0

I think the answer is G=/2.(8) =(G=4)

svlad2 [7]3 years ago

3 0

$\bf \cfrac{2}{x}=\cfrac{x}{8}\implies 16=x^2\implies \sqrt{16}=x\implies 4=x$

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Which value is equivalent to sin 42?

mafiozo [28]

ANSWER-COS 48 DEGREE

4 0

3 years ago

The polygon Q(3, 2), R(6,5), S(6, 2)

ycow [4]

Answer:

Q' (12,8) , R'( 24,20) and S' (24,8)

Step-by-step explanation:

Here, we want to get the coordinates of the image after dilating the pre-image by a scale factor of 4

What we have to do here is to multiply each of the coordinate on the pre-image by 4

We have this as;

Q' = (4*3, 2 * 4) = (12,8)

R' = (4*6, 4*5) = (24,20)

s' = (4*6, 4*2) = (24,8)

4 0

3 years ago

Solution for dy/dx+xsin 2 y=x^3 cos^2y

vichka [17]

Rearrange the ODE as

$\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y=x^3\cos^2y$
$\sec^2y\dfrac{\mathrm dy}{\mathrm dx}+x\sin2y\sec^2y=x^3$

Take $u=\tan y$ , so that $\dfrac{\mathrm du}{\mathrm dx}=\sec^2y\dfrac{\mathrm dy}{\mathrm dx}$ .

Supposing that $|y|$ , we have $\tan^{-1}u=y$ , from which it follows that

$\sin2y=2\sin y\cos y=2\dfrac u{\sqrt{u^2+1}}\dfrac1{\sqrt{u^2+1}}=\dfrac{2u}{u^2+1}$
$\sec^2y=1+\tan^2y=1+u^2$

So we can write the ODE as

$\dfrac{\mathrm du}{\mathrm dx}+2xu=x^3$

which is linear in $u$ . Multiplying both sides by $e^{x^2}$ , we have

$e^{x^2}\dfrac{\mathrm du}{\mathrm dx}+2xe^{x^2}u=x^3e^{x^2}$
$\dfrac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]=x^3e^{x^2}$

Integrate both sides with respect to $x$ :

$\displaystyle\int\frac{\mathrm d}{\mathrm dx}\bigg[e^{x^2}u\bigg]\,\mathrm dx=\int x^3e^{x^2}\,\mathrm dx$
$e^{x^2}u=\displaystyle\int x^3e^{x^2}\,\mathrm dx$

Substitute $t=x^2$ , so that $\mathrm dt=2x\,\mathrm dx$ . Then

$\displaystyle\int x^3e^{x^2}\,\mathrm dx=\frac12\int 2xx^2e^{x^2}\,\mathrm dx=\frac12\int te^t\,\mathrm dt$

Integrate the right hand side by parts using

$f=t\implies\mathrm df=\mathrm dt$
$\mathrm dg=e^t\,\mathrm dt\implies g=e^t$
$\displaystyle\frac12\int te^t\,\mathrm dt=\frac12\left(te^t-\int e^t\,\mathrm dt\right)$

You should end up with

$e^{x^2}u=\dfrac12e^{x^2}(x^2-1)+C$
$u=\dfrac{x^2-1}2+Ce^{-x^2}$
$\tan y=\dfrac{x^2-1}2+Ce^{-x^2}$

and provided that we restrict $|y|$ , we can write

$y=\tan^{-1}\left(\dfrac{x^2-1}2+Ce^{-x^2}\right)$

5 0

3 years ago

Can someone help me with these questions??

Naddika [18.5K]

What questions I mean there is no questions

4 0

2 years ago

2. If AB = 4x + 9. BC = 5x + 2, and AC = 56, then find the value for . AB, BC.<br> A<br> B

inysia [295]

Answer:

AB=29; BC=27

Step-by-step explanation:

So they told us AB=4x+9 and that BC=5x+2, and AC=56 , now to help with the question you can draw this information on a number line. Now on a number you can see that basically AC=AB+BC.

So you would write it as such,,

4x+9+5x+2=56

Combine like terms

9x+11=56

Now you have to isolate the x by itself but first get rid of the 11.

9x+11-11=56-11

You would get

9x=45

Here you can divide 9 by both sides to isolate x.

9x/9=45/9

{x=5}

Now to find the value for both substitue x in the equations for both

1. AB=4x+9 where x is 5

4(5)+9 =AB

20+9 =AB

29=AB

You would do the same with BC

2. BC= 5x+2 where x is 5

5(5)+2= BC

25+2= BC

27=BC

If you want to check your answers you can just substitute x for 5 in the first equation we did where AC=AB+BC

6 0

3 years ago