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lord [1]
3 years ago
13

Translate quadrilateral ABCD 5 units to the right and 6 units down.

Mathematics
1 answer:
dezoksy [38]3 years ago
6 0

Answer:

D. A'(1,-7), B'(2,-1), C'8,-2) and D'(7,-5)

Step-by-step explanation:

Probably too late, but I took the quiz

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use a calculator / solve for x calculator

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A candle in the shape of a circular cone has a base of radius r and a height of h that is the same length as the radius. which e
ladessa [460]

Answer:

\frac{r(1-\sqrt{2})}{-3}

Step-by-step explanation:

Volume of cone = \frac{1}{3} \pi r^{2} h

Since we are given that a circular cone has a base of radius r and a height of h that is the same length as the radius

                          = \frac{1}{3} \pi r^{2} \times r

                          = \frac{1}{3} \pi r^{3}

Surface area of cone including 1 base = \pi r^{2} +\pi\times r \times \sqrt{r^2+h^2}

Since r = h

So, area = \pi r^{2} +\pi\times r \times \sqrt{r^2+r^2}

              = \pi r^{2} +\pi\times r \times \sqrt{2r^2}

              = \pi r^{2} +\pi\times r^2 \times \sqrt{2}

Ratio of volume of cone to its surface area including base :

\frac{\frac{1}{3} \pi r^{3}}{\pi r^{2} +\pi\times r^2 \times \sqrt{2}}

\frac{\frac{1}{3}r}{1+\sqrt{2}}

\frac{r}{3(1+\sqrt{2})}

Rationalizing

\frac{r}{3(1+\sqrt{2})} \times \frac{1-\sqrt{2}}{1-\sqrt{2}}

\frac{r(1-\sqrt{2})}{-3}

Hence the ratio the ratio of the volume of the candle to its surface area(including the base) is \frac{r(1-\sqrt{2})}{-3}

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