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Ugo [173]
3 years ago
15

Help!! Will give brainliest

Mathematics
2 answers:
patriot [66]3 years ago
4 0

Answer:

c

Step-by-step explanation:

Arturiano [62]3 years ago
3 0

c

................................................................

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What is the area of this trapezoid?<br><br><br><br> Enter your answer in the box.<br><br> units2
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Answer:

area=88 square units

Step-by-step explanation:

area of trapezoid:(sum of the two bases)/2 ×h

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Drag the tiles to the correct boxes to complete the pairs.
Advocard [28]

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did you answer this already

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Miguel wants to bike a total of 100 miles over the course of one week in the summer. He logs how much he bikes each day in the t
bagirrra123 [75]

Answer:

Change all of the number into the same form for ease of adding.

16.3 + 14.5 + 18.75 + 12.25 + 22.6 = 84.4

100 - 84.4 = 15.6

Miguel has to ride a total of 15.6 miles on Saturday and Sunday

4 0
4 years ago
Please help! I will mark you as brainliest!
Lisa [10]

Answer:

105.6 cm²

Step-by-step explanation:

Area of full circle = π × r²

Semicircle = Half it

3.142 × 8.2² = 211.26808 ( Full circle)

211.26808 ÷ 2 = 105.63404

105.6 cm²

8 0
3 years ago
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Find the volume v of the described solid s. the base of s is an elliptical region with boundary curve 4x2 + 9y2 = 36. cross-sect
Tasya [4]
4x^2+9y^2=36\iff\dfrac{x^2}9+\dfrac{y^2}4=1

defines an ellipse centered at (0,0) with semi-major axis length 3 and semi-minor axis length 2. The semi-major axis lies on the x-axis. So if cross sections are taken perpendicular to the x-axis, any such triangular section will have a base that is determined by the vertical distance between the lower and upper halves of the ellipse. That is, any cross section taken at x=x_0 will have a base of length

\dfrac{x^2}9+\dfrac{y^2}4=1\implies y=\pm\dfrac23\sqrt{9-x^2}
\implies \text{base}=\dfrac23\sqrt{9-{x_0}^2}-\left(-\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac43\sqrt{9-{x_0}^2}

I've attached a graphic of what a sample section would look like.

Any such isosceles triangle will have a hypotenuse that occurs in a \sqrt2:1 ratio with either of the remaining legs. So if the hypotenuse is \dfrac43\sqrt{9-{x_0}^2}, then either leg will have length \dfrac4{3\sqrt2}\sqrt{9-{x_0}^2}.

Now the legs form a similar triangle with the height of the triangle, where the legs of the larger triangle section are the hypotenuses and the height is one of the legs. This means the height of the triangular section is \dfrac4{3(\sqrt2)^2}\sqrt{9-{x_0}^2}=\dfrac23\sqrt{9-{x_0}^2}.

Finally, x_0 can be chosen from any value in -3\le x_0\le3. We're now ready to set up the integral to find the volume of the solid. The volume is the sum of the infinitely many triangular sections' areas, which are

\dfrac12\left(\dfrac43\sqrt{9-{x_0}^2}\right)\left(\dfrac23\sqrt{9-{x_0}^2}\right)=\dfrac49(9-{x_0}^2)

and so the volume would be

\displaystyle\int_{x=-3}^{x=3}\frac49(9-x^2)\,\mathrm dx
=\left(4x-\dfrac4{27}x^3\right)\bigg|_{x=-3}^{x=3}
=16

6 0
3 years ago
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