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Sonbull [250]
3 years ago
8

What is the value between the 9s in 709,946,107

Mathematics
2 answers:
kodGreya [7K]3 years ago
6 0
9 million and 9 hundred thousand
Vitek1552 [10]3 years ago
4 0
I think the answer is 8,100,000
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The staff of the Westport public library signed up 6 volunteers in the last 21 days. If the library staff continues to sign up v
Andre45 [30]
Both 6 and 21 are divisive by 3 so each 7 day week the library hires 2 people.
answer is 2
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3 years ago
50 POINTS What is the greatest common factor of 4k, 18k4, and 12?
vredina [299]

Answer:

The greatest common factor of 4k, 18k4, and 12 is 2.

Answer A 2

Hope this helps

6 0
2 years ago
Which of the following quadrilaterals are classified as types of parallelograms?
Natali5045456 [20]
Square, Rhombus and rectangle.
8 0
3 years ago
Read 2 more answers
These liness pass through the same point
anyanavicka [17]

Answer:

All of them except B

Step-by-step explanation:

Bisecting is splitting a line in half

Parallel lines never touch each other

Intersecting lines intersect

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8 0
2 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
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