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enot [183]
3 years ago
5

Help please And thank you

Mathematics
2 answers:
Solnce55 [7]3 years ago
5 0

Answer:

Step-by-step explanation:

kari74 [83]3 years ago
4 0

Answer:

cant see image

Step-by-step explanation:

.

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Help 15 points!!!!!!!!!!!!!!!!!!!!!
Svetllana [295]

Answer:

see explanation

Step-by-step explanation:

Using the tangent ratio in the right triangle

tan A = \frac{opposite}{adjacent} = \frac{BC}{AC} = \frac{7}{8} , then

∠ A = tan^{-1} (\frac{7}{8} ) ≈ 41° ( to the nearest degree )

The sum of the angles in the triangle = 180° , then

∠ B + 41° + 90° = 180°

∠ B + 131° = 180° ( subtract 131° from both sides )

∠ B = 49°

Using Pythagoras' identity in the right triangle

AB² = BC² + AC² = 7² + 8² = 49 + 64 = 113 ( take square root of both sides )

AB = \sqrt{113} ≈ 10.6 ( to the nearest tenth )

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The ratio of the circumferences of two circles is 2:3. If the large circle has a radius of 39cm what is the radius of the small
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Read 2 more answers
Both the La Plata river dolphin (Pontoporia blainvillei) and
Citrus2011 [14]

Answer:

<em>1. A sperm whale is </em><em>3</em><em> </em><em>orders of magnitude</em><em> heavier than a La Plata river dolphin; 2. The radius of the larger ball is </em><em>one (1) order of magnitude</em><em> bigger than the radius of the smaller ball; 3. The volume of the larger ball is </em><em>3 orders of magnitude</em><em> bigger than the volume of the smaller ball</em>.

Step-by-step explanation:

If we expressed a number as:

\\ N = a * 10^{b} (1)

Where

\\ \frac{1}{\sqrt{10}} \leq a < \sqrt{10} (2)

or

\\ 1 \leq a < 10 (3)

Then, <em>b</em> represents the <em>order of magnitude </em>of such a number (<em>Order of magnitude (2020), </em>in Wikipedia).

The order of magnitude can be defined as "...the smallest power of ten needed to represent a quantity" (Weisstein, Eric W. "Order of Magnitude". From MathWorld--A Wolfram Web Resource).

Having gathered all this information, we can proceed as follows:

<h3>First case</h3>

The<em> La Plata river dolphin</em> weighs between 30 and 50kg and the sperm whale weighs between 35,000 and 40,000kg.

Then, considering (1) and (3) to express the dolphin and whale's weight (since in this way the order of magnitude is the same as the exponent part in the <em>scientific notation</em>):

\\ 30kg \leq Dolphin_{weight} \leq 50kg

\\ 3*10^{1}kg \leq Dolphin_{weight} \leq 5*10^{1}kg

\\ 35000kg \leq Whale_{weight} \leq 40000kg

\\ 3.5*10^{4}kg \leq Whale_{weight} \leq 4.0*10^{4}kg

Since the range for the weights are in the same order of magnitude for both dolphin and whale (considering the definition above):

\\ Dolphin_{weight} = 10^{1}\;(order\;of\;magnitude=1)

\\ Whale_{weight} = 10^{4}\;(order\;of\;magnitude=4)

Then

\\ \frac{Whale_{weight} = 10^{4}}{Dolphin_{weight} = 10^{1}}

\\ \frac{Whale_{weight}}{Dolphin_{weight}} = \frac{10^{4}}{10^{1}}

\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{4-1}

\\ \frac{Whale_{weight}}{Dolphin_{weight}} = 10^{3}

Thus

<em>A sperm whale is </em><em>3</em><em> </em><em>orders of magnitude</em><em> heavier than a La Plata river dolphin.</em>

<h3>Second case</h3>

Following the same reasoning, we can conclude that <em>the radius of the larger ball is </em><em>one (1) order of magnitude</em><em> bigger than the radius of the smaller ball:</em>

\\ \frac{Larger\;ball_{radius}}{Smaller\;ball_{radius}} = \frac{10^{1}}{10^{0}}

\\ \frac{Larger\;ball_{radius}}{Smaller\;ball_{radius}} = 10^{1-0} = 10^{1}

<h3>Third case</h3>

For this case, we need to calculate <em>the volume of a sphere</em> for both radii (1cm and 10cm).

The volume of a sphere is

\\ V_{sphere} = \frac{4}{3}*\pi*R^{3}

Then, the volume of the <em>ball of radius 1cm</em> is:

\\ V_{radius=1} = \frac{4}{3}*\pi*(1cm)^{3}

\\ V_{radius=1} \approx 4.19*10^{0}cm^{3}

And, the volume of the <em>ball of radius 10cm</em> is:

\\ V_{radius=10} = \frac{4}{3}*\pi*(10cm)^{3}

\\ V_{radius=10} \approx 4.19*10^{3}cm^{3}

Thus

\\ \frac{10^{3}}{10^{0}} = 10^{3}

As a result, <em>the volume of the larger ball is </em><em>3 orders of magnitude</em><em> bigger than the volume of the smaller ball</em>.

4 0
3 years ago
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