Which quadrilateral will always have 4- fold reflectional symmetry

Mathematics

1 answer:

Talja [164]3 years ago

5 0

I'm pretty sure it's the square. The square can be divided into 4 equal parts, each part able to reflect each other.

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Find the length of the radius of the following circle. (x+2)^2+y^2=10

Leokris [45]

<span>(x - h)² + (y - k)² = r²

is the equation of the circle with center (h, k) and radius r units.

Thus (x + 2)² + y² = 10 is the equation of a circle whose center is at the point (-2, 0) and having a radius of √10 units<u>
</u></span>

6 0

3 years ago

75 points. Will give certified if work is shown and the answer is correct.

Svetach [21]

Volume of the first dwarf planet (r₁ = 832 mi):

$V_1=\dfrac{4}{3}\cdot\pi\cdot r_1^3=\dfrac{4}{3}\cdot\pi\cdot 832^3=\dfrac{2303721472}{3}\pi\approx7.679\cdot10^8\pi\,\text{mi}^3$

Volume of the second dwarf planet (r₂ = 829 mi):

$V_2=\dfrac{4}{3}\cdot\pi\cdot r_2^3=\dfrac{4}{3}\cdot\pi\cdot 829^3=\dfrac{2278891156}{3}\pi\approx7.5963\cdot10^8\pi\,\text{mi}^3$

So difference between the volumes is:

$V_1-V_2\approx7.679\cdot10^8\pi-7.5963\cdot10^8\pi=0.0827\cdot10^8\pi=\boxed{8270000\pi\,\text{mi}^3}$

or if we want exact value (we use (a³-b³) = (a-b)(a²+ab+b²) ):

$V_1-V_2=\dfrac{4}{3}\cdot\pi\cdot r_1^3-\dfrac{4}{3}\cdot\pi\cdot r_2^3=\dfrac{4}{3}\pi(r_1^3-r_2^3)=\dfrac{4}{3}\pi(832^3-829^3)=\\\\\\=\dfrac{4}{3}\pi(832-829)(832^2+832\cdot829+829^2)=\\\\\\=\dfrac{4}{3}\pi\cdot3(692224+689728+687241)=4\pi\cdot2069193=\boxed{8276772\pi\,\text{mi}^3}$

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3 years ago

An envelope from the post office is 3 inches wide with a total area of 30 square inches.what is the height of the envelope

ad-work [718]

The height of the envelope is 10. This is because you divide the area by 3. So, the missing side is 10 inches.

7 0

3 years ago

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In ΔBCD, the measure of ∠D=90°, the measure of ∠B=31°, and CD = 1.6 feet. Find the length of BC to the nearest tenth of a foot.

son4ous [18]

Answer:

3.1066≈3.1 feet