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Nana76 [90]
3 years ago
10

Which compound sentences is not correct punctuated

Mathematics
1 answer:
vlabodo [156]3 years ago
6 0
The answer to this question is the third one down
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*100 points* A circle has a diameter with endpoints (-7, -1) and (-5, -9).
earnstyle [38]

Diameter

  • √(-7+5)²+(-1+9)²
  • √2²+8²
  • √4+64
  • √68
  • 2√17

Radius

  • 2√17/2
  • √17units

Now

Centre

(h,k)

  • (-7-5/2,-1-9/2)
  • (-12/2,-10/2)
  • (-6,-5)

Equation

  • (x-h)²+(y-k)²=r²
  • (x+6)²+(y+5)²=(√17)²
  • (x+6)²+(y+5)²=17

Option A

8 0
2 years ago
Read 2 more answers
0.34 to the lowest term
My name is Ann [436]
0.34 =  \dfrac{34}{100}  =  \dfrac{17 \times 2}{50 \times 2} =  \dfrac{17}{50}
5 0
3 years ago
Eileen picks up 5 movies at a movie store and calculates that it will cost her $32.50 before tax. A customer in front of Eileen
Svetradugi [14.3K]

Answer:

$34.125

Step-by-step explanation:

5% of 32.50 is 1.625

1.625+32.50=34.125

$34.125

6 0
3 years ago
Read 2 more answers
Please help w this! Its a calculus question! look at the picture for the problem,
neonofarm [45]

Since you mentioned calculus, perhaps you're supposed to find the area by integration.

The square is circumscribed by a circle of radius 6, so its diagonal (equal to the diameter) has length 12. The lengths of a square's side and its diagonal occur in a ratio of 1 to sqrt(2), so the square has side length 6sqrt(2). This means its sides occur on the lines x=\pm3\sqrt2 and y=\pm3\sqrt2.

Let R be the region bounded by the line x=3\sqrt2 and the circle x^2+y^2=36 (the rightmost blue region). The right side of the circle can be expressed in terms of x as a function of y:

x^2+y^2=36\implies x=\sqrt{36-y^2}

Then the area of this circular segment is

\displaystyle\iint_R\mathrm dA=\int_{-3\sqrt2}^{3\sqrt2}\int_{3\sqrt2}^{\sqrt{36-y^2}}\,\mathrm dx\,\mathrm dy

=\displaystyle\int_{-3\sqrt2}^{3\sqrt2}(\sqrt{36-y^2}-3\sqrt2)\,\mathrm dy

Substitute y=6\sin t, so that \mathrm dy=6\cos t\,\mathrm dt

=\displaystyle\int_{-\pi/4}^{\pi/4}6\cos t(\sqrt{36-(6\sin t)^2}-3\sqrt2)\,\mathrm dt

=\displaystyle\int_{-\pi/4}^{\pi/4}(36\cos^2t-18\sqrt2\cos t)\,\mathrm dt=9\pi-18

Then the area of the entire blue region is 4 times this, a total of \boxed{36\pi-72}.

Alternatively, you can compute the area of R in polar coordinates. The line x=3\sqrt2 becomes r=3\sqrt2\sec\theta, while the circle is given by r=6. The two curves intersect at \theta=\pm\dfrac\pi4, so that

\displaystyle\iint_R\mathrm dA=\int_{-\pi/4}^{\pi/4}\int_{3\sqrt2\sec\theta}^6r\,\mathrm dr\,\mathrm d\theta

=\displaystyle\frac12\int_{-\pi/4}^{\pi/4}(36-18\sec^2\theta)\,\mathrm d\theta=9\pi-18

so again the total area would be 36\pi-72.

Or you can omit using calculus altogether and rely on some basic geometric facts. The region R is a circular segment subtended by a central angle of \dfrac\pi2 radians. Then its area is

\dfrac{6^2\left(\frac\pi2-\sin\frac\pi2\right)}2=9\pi-18

so the total area is, once again, 36\pi-72.

An even simpler way is to subtract the area of the square from the area of the circle.

\pi6^2-(6\sqrt2)^2=36\pi-72

6 0
3 years ago
How to divide 20 by radical 3
Alex Ar [27]

Answer:

"How do you divide a whole number by a radical?

Method 1 Dividing Radicands

Set up a fraction. If your expression is not already set up like a fraction, rewrite it this way. ...

Use one radical sign. If your problem has a square root in the numerator and denominator, you can place both radicands under one radical sign. ...

Divide the radicands. ...

Simplify, if necessary.

4 Simple and Easy Ways to Divide Square Roots - wikiHow

https://m.wikihow.com/Divide-Square-Roots "


Step-by-step explanation:


5 0
3 years ago
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