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Darya [45]
2 years ago
6

Which fraction equals 4?

Mathematics
1 answer:
Ilya [14]2 years ago
7 0

Answer: 8/2

Step-by-step explanation:

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Please help it will help a lot i really need this!!
Lorico [155]

Answer:

A' = (0,0)

B'=(8,0)

C'=(8,2)

D'=(0,2)

It is not a rigid motion.

Step-by-step explanation:

To use the mapping rule, substitute the original x and y values in it.

The coordinates of A are (0,0).

Using the mapping rule, the x coordinate of A' = 2x0 = 0

Using the mapping rule, the y coordinate of A' = (1/2)x0=0

So A' will not change locations. The image will be at (0,0).

The coordinates of B are (4,0)

Using the mapping rule, the x-coordinate of B' = 2x4 = 8

Using the mapping rule, the y-coordinate of B' = (1/2)x0=0

Therefore the image of B' will be located at coorindate (8,0)

The coorindates of C are (4,4).

Using the mapping rule, the x-coordinate of C' = 2x4=8

Using the mapping rule, the y-coordinate of C' = (1/2)x4=2

Therefore the image of C' will be located at coordinate (8,2)

The coordinates of D are (0,4)

Using the mapping rule, the x-coordinate of D' = 2x0 = 0

Using the mapping rule, the y-coordinate of D' = (1/2)x4=2

Therefore the image of D' will be located at coordinate (0,2)

<em>Is the transformation a rigid motion?</em>

NO, this transformation is not a rigid motion because the relative distance between the points does not stay the same after they have been transformed. The transformation is not a translation , rotation, reflection, nor glide reflection.

6 0
2 years ago
Will give brainliest to the homie that gets it right! please help!
12345 [234]
I'm pretty sure the answer to your question is -3/5 + 4/5. C.

4 0
3 years ago
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
Find the amount of money a woman owes at the end of 3 years if 5% interest is compounded continuously on her $2500 debt
stepladder [879]

Answer:

$ 2904.59

Step-by-step explanation:

For CONTINUOUS compounding    FV = PV e^it

FV = future value    PV = present value   i = decimal interest   t = years

FV = 2500 e^(.05 * 3) = 2904.59

5 0
2 years ago
A rectangle garden has a length of 12 feet. you need 36 feet of fencing to enclose the garden. what is the width of the garden ?
iren [92.7K]

Let us suppose the width is x feet

the length is given 12 feet

Perimeter of teh garden = 2( Length + width )= 2( 12+ x)

Fencing required = perimeter of garden

36 = 2( 12+x)

36 = 2*12 + 2x

36 = 24 + 2x

36-24 = 2x

12 = 2x

12/2 = x

6 = x

Width = x feet = 6 feet .Answer .

7 0
3 years ago
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