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3 years ago

Someone help please?! Need right answers only don’t come for the points will mark brainliest

Mathematics

1 answer:

Kaylis [27]3 years ago

6 0

Hi there!

These can probably be done on your own. You just gotta know what to do! :)

Let's take #1 for example. You (or maybe a classmate/teacher showed you?) plotted the points. Mark each point with the given letter, so you don't get lost. Then, you reflected it over the y-axis.

Think of it as a mirror. Say you held a picture of a rhombus up to it. You would see the rhombus, yourself, and whatever was in the background reflected back at you. You step closer, the image steps closer. You turn the rhombus, and the image also turns. This principle can be used here!

So, keep doing what you're doing. Here's a step-by-step:

1.) Plot each point, and mark its name. For example, 'B' is (-6,7), and you write 'B' next to the point.

2.) Double check the point are exactly where they need to be

3.) Connect each point with a straight line. You can use a ruler, student ID, whatever as a straightedge, but it looks neater

4.) Draw a line for the axis. For example, if y=0, draw a straight line again there. (hint: that's the y-axis!)

5.) Double check that everything is right so far again. This is easy to mess up!

6.) Reflect each point over the axis. Another example, (-3, 2) becomes (3, 2). Mark this with an apostrophe (') to signal the point as prime, or the reflected point. For example, B becomes B' (B prime)

7.) Check one final time

If you found this especially helpful, I'd appreciate if you'd vote me Brainliest for your answer. I want to be able to assist more users one-on-one! :)

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

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)

$\\ 1 \leq a < 10$ (3)

Then, b represents the order of magnitude of such a number (Order of magnitude (2020), 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 La Plata river dolphin 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 scientific notation):

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

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

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

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

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

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

<h3>Second case</h3>

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

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

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

<h3>Third case</h3>

For this case, we need to calculate the volume of a sphere 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 ball of radius 1cm is:

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

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

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

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

Thus

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

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

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