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navik [9.2K]
3 years ago
10

Round 8.96278650871 to the nearest millionth.

Mathematics
2 answers:
QveST [7]3 years ago
7 0
The nearest millionth would be 8.962787 since 5 and up means you round up and so the 6 you would round up to a 7
nordsb [41]3 years ago
4 0

Answer:

The answer is 8.96278<u>7</u> .

Step-by-step explanation:

The number above is :

8 = Ones place

8.9 = Tenth place

8.96 = Hundredth place

8.962 = Thousandth place

8.9627 = Ten thousandth place

8.96278 = Hundred thousandth place

8.962786 = Millionth place

8.96278<u>7</u> = Nearest millionth

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Equation of the line that is perpendicular to y= 1/8x -2 and goes through the point <br> (-2,-3)
larisa [96]
The answer to this equation is y=-8x-19



I hope this helps!!!!!!!!
7 0
3 years ago
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Two points in the Cartesian plane are A(2.00 m, −4.00 m) and B(−3.00 m, 3.00 m). Find the distance between them and their polar
Brrunno [24]

The distance between the two points is d=8.6m

The polar coordinate of A is \left(4.47,296.57\right)

The polar coordinate of B is \left(4.24,135\right)

Explanation:

The two points are A(2,-4) and B(-3,3)

The distance between two points is given by,

d=\sqrt{(2+3)^{2}+(-4-3)^{2}}\\d=\sqrt{(5)^{2}+(-7)^{2}}\\d=\sqrt{25+49}\\d=8.6

Thus, the distance between the two points is d=8.6m

The polar coordinates of A can be written as (Distance, tan^{-1} \frac{y}{x} )

Distance = \sqrt{x^{2} +y^{2} }

Substituting A(2,-4), we get,

Distance = \sqrt{2^{2} +(-4)^{2} }=\sqrt{4+16 }=4.47

tan^{-1} \frac{y}{x} =tan^{-1} \frac{-4}{2}=-63.43

To make the angle positive, let us add 360,

\theta=360-63.43=296.57

The polar coordinate of A is \left(4.47,296.57\right)

Similarly, The polar coordinate of B can be written as (Distance, tan^{-1} \frac{y}{x} )

Distance = \sqrt{x^{2} +y^{2} }

Substituting B(-3,3), we get,

Distance = \sqrt{(3)^{2}+(3)^{2}}=4.24

tan^{-1} \frac{y}{x} =tan^{-1} \frac{3}{-3}=-45

To make the angle positive, let us add 360,

\theta=180-45=135^{\circ}

The polar coordinate of B is \left(4.24,135\right)

5 0
3 years ago
Find the surface area of the part of the sphere x2+y2+z2=81 that lies above the cone z=x2+y2−−−−−−√
TiliK225 [7]
Parameterize the part of the sphere by

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with 0\le u\le2\pi and 0\le v\le\dfrac\pi4. Then

\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|=81\sin v\,\mathrm du\,\mathrm dv

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\displaystyle\iint_S\mathrm dS=81\int_{v=0}^{v=\pi/4}\int_{u=0}^{u=2\pi}\sin v\,\mathrm du\,\mathrm dv=81(2\pi)\left(1-\dfrac1{\sqrt2}\right)=(162-81\sqrt2)\pi
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I need help plss!! I don’t understand
MA_775_DIABLO [31]

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3 years ago
What does the tape measure say Measurement # 2 is?
leonid [27]

Answer:

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Step-by-step explanation:

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Call and ask for pinus step by step

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3 years ago
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