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snow_tiger [21]
2 years ago
13

Find the volume of a sphere with a surface area of 2916 squared pi

Mathematics
1 answer:
Svet_ta [14]2 years ago
6 0

Answer:

it would be somewhere around 29161231

Step-by-step explanation:

it is m8

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-3(6 - 3k)<br>I need to Simplify the expression​
sergij07 [2.7K]

Answer:

-18+9k

Step-by-step explanation:

If you use the distributive property, you get:

(-3*6) and (-3*-3k)

If you simplify each of them you get:

-18+9k

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3 years ago
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Algebra help! Please help me!
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Your answer would most likely be C. 2^1/3 i'm sorry if I got it wrong
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Henrietta has a sample of a pure substance in a beaker. She heats the substance to a temperature of 100°C. What will determine i
erica [24]
The answer is c
have a good day
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2 years ago
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Looking at the top of tower A and base of tower B from points C and D, we find that ∠ACD = 60°, ∠ADC = 75° and ∠ADB = 30°. Let t
katrin2010 [14]

Answer:

\text{Exact: }AB=25\sqrt{6},\\\text{Rounded: }AB\approx 61.24

Step-by-step explanation:

We can use the Law of Sines to find segment AD, which happens to be a leg of \triangle ACD and the hypotenuse of \triangle ADB.

The Law of Sines states that the ratio of any angle of a triangle and its opposite side is maintained through the triangle:

\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}

Since we're given the length of CD, we want to find the measure of the angle opposite to CD, which is \angle CAD. The sum of the interior angles in a triangle is equal to 180 degrees. Thus, we have:

\angle CAD+\angle ACD+\angle CDA=180^{\circ},\\\angle CAD+60^{\circ}+75^{\circ}=180^{\circ},\\\angle CAD=180^{\circ}-75^{\circ}-60^{\circ},\\\angle CAD=45^{\circ}

Now use this value in the Law of Sines to find AD:

\frac{AD}{\sin 60^{\circ}}=\frac{100}{\sin 45^{\circ}},\\\\AD=\sin 60^{\circ}\cdot \frac{100}{\sin 45^{\circ}}

Recall that \sin 45^{\circ}=\frac{\sqrt{2}}{2} and \sin 60^{\circ}=\frac{\sqrt{3}}{2}:

AD=\frac{\frac{\sqrt{3}}{2}\cdot 100}{\frac{\sqrt{2}}{2}},\\\\AD=\frac{50\sqrt{3}}{\frac{\sqrt{2}}{2}},\\\\AD=50\sqrt{3}\cdot \frac{2}{\sqrt{2}},\\\\AD=\frac{100\sqrt{3}}{\sqrt{2}}\cdot\frac{ \sqrt{2}}{\sqrt{2}}=\frac{100\sqrt{6}}{2}={50\sqrt{6}}

Now that we have the length of AD, we can find the length of AB. The right triangle \triangle ADB is a 30-60-90 triangle. In all 30-60-90 triangles, the side lengths are in the ratio x:x\sqrt{3}:2x, where x is the side opposite to the 30 degree angle and 2x is the length of the hypotenuse.

Since AD is the hypotenuse, it must represent 2x in this ratio and since AB is the side opposite to the 30 degree angle, it must represent x in this ratio (Derive from basic trig for a right triangle and \sin 30^{\circ}=\frac{1}{2}).

Therefore, AB must be exactly half of AD:

AB=\frac{1}{2}AD,\\AB=\frac{1}{2}\cdot 50\sqrt{6},\\AB=\frac{50\sqrt{6}}{2}=\boxed{25\sqrt{6}}\approx 61.24

3 0
3 years ago
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Which ratio froms a proportional with 16/20? A 4/10 B 32/50 C 8/12 D 4/5​
kiruha [24]
D . 4/5 16/4 = 4 | 20/4=5.
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