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Brums [2.3K]
2 years ago
9

In which triangle is the measure of the unknown angle, x, equal to the value of sin–1(StartFraction 5 Over 8. 3 EndFraction)?.

Mathematics
1 answer:
almond37 [142]2 years ago
6 0

The triangle is a right-angled triangle with a perpendicular side of 5 and a hypotenuse of 8.3.

<h2 /><h2>Given to us</h2>

\theta = sin^{-1}\dfrac{5}{8.3}

<h2>Sine θ</h2>

as we know that sine for any angle is the ratio of its perpendicular to the hypotenuse,  and trigonometry is only applicable to a right-angled triangle.

therefore,

\theta = sin^{-1}\dfrac{5}{8.3}\\\\sin \theta=\dfrac{5}{8.3}=\dfrac{Perpendicular}{Hypotenuse}\\

Hence, the triangle is a right-angled triangle with a perpendicular side of 5 and a hypotenuse of 8.3.

Learn more about Sine:

brainly.com/question/18055768

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Yes
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In the illustration below, the three cube-shaped tanks are identical. The spheres in any given tank
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Answer:

1) Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) Amount \ of \, water \ remaining \ in \, the \ tank \ is \  \frac{x^3(6-\pi) }{6}

Step-by-step explanation:

1) Here we have;

First tank A

Volume of tank = x³

The  volume of the sphere = \frac{4}{3} \pi r^3

However, the diameter of the sphere = x therefore;

r = x/2 and the volume of the sphere is thus;

volume of the sphere = \frac{4}{3} \pi \frac{x^3}{8}= \frac{1}{6} \pi x^3

For tank B

Volume of tank = x³

The  volume of the spheres = 8 \times \frac{4}{3} \pi r^3

However, the diameter of the spheres 2·D = x therefore;

r = x/4 and the volume of the sphere is thus;

volume of the spheres = 8 \times \frac{4}{3} \pi (\frac{x}{4})^3= \frac{x^3 \times \pi }{6}

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Volume of tank = x³

The  volume of the spheres = 64 \times \frac{4}{3} \pi r^3

However, the diameter of the spheres 4·D = x therefore;

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volume of the spheres = 64 \times \frac{4}{3} \pi (\frac{x}{8})^3= \frac{x^3 \times \pi }{6}

Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) For the 4th tank, we have;

number of spheres on side of the tank, n is given thus;

n³ = 512

∴ n = ∛512 = 8

Hence we have;

Volume of tank = x³

The  volume of the spheres = 512 \times \frac{4}{3} \pi r^3

However, the diameter of the spheres 8·D = x therefore;

r = x/16 and the volume of the sphere is thus;

volume of the spheres = 512\times \frac{4}{3} \pi (\frac{x}{16})^3= \frac{x^3 \times \pi }{6}

Amount of water remaining in the tank is given by the following expression;

Amount of water remaining in the tank = Volume of tank - volume of spheres

Amount of water remaining in the tank = x^3 - \frac{x^3 \times \pi }{6} = \frac{x^3(6-\pi) }{6}

Amount \ of \ water \, remaining \, in \, the \ tank =  \frac{x^3(6-\pi) }{6}.

5 0
3 years ago
What is 3/4 + 2/3 ???
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To add fractions, first convert them to have equal denominators.

3/4= 9/12 (multiply both sides by 3)
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From here, add the numerators while keeping the denominator.
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