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JulsSmile [24]
4 years ago
7

Is the following relation a function?

Mathematics
2 answers:
iVinArrow [24]4 years ago
4 0
Yes, it is definitely a function. Although one might say that it is not a one to one or injective function because of the presence of (<span>-1,-2) and (6,-2).. The correct option among the two options that are given in the question is the second option. I hope that this is the answer that has actually come to your desired help.</span>
Nady [450]4 years ago
3 0

Answer:Yes, it is definitely a function. Although one might say that it is not a one to one or injective function because of the presence of (-1,-2) and (6,-2).. The correct option among the two options that are given in the question is the second option. I hope that this is the answer that has actually come to your desired help

Step-by-step explanation: I took the test on FLVS

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Trevon’s school is selling tickets to a fall musical. On the first day of ticket sales the school sold 13 adult tickets and 12 s
julsineya [31]

9514 1404 393

Answer:

  • adult: $7
  • student: $10

Step-by-step explanation:

Let a and s represent the prices of adult and student tickets, respectively.

  13a +12s = 211 . . . . . . ticket sales the first day

  5a +3s = 65 . . . . . . . ticket sales the second day

Subtracting the first equation from 4 times the second gives ...

  4(5a +3s) -(13a +12s) = 4(65) -(211)

  7a = 49 . . . . . . . simplify

  a = 7 . . . . . . . divide by 7

  5(7) +3s = 65 . . . . substitute into the second equation

  3s = 30 . . . . . . . subtract 35

  s = 10 . . . . . . . divide by 3

The price of one adult ticket is $7; the price of one student ticket is $10.

7 0
3 years ago
How many 3/6 are there in 1 whole <br><br><br> Don’t give me glibberish answers or I will report u
Aliun [14]

Answer:2

Step-by-step explanation:because

x=1x6/3

X=2

Its not gibberish so gimme brainliest

5 0
4 years ago
Read 2 more answers
There are five cars in line at a stoplight. Each of these cars is a different color and different type of car. Diego is driving
adell [148]
Ok, here we go:
5 cars total.
1. Rachel - first car in line, yellow, SUV
2. blue car (Rachel is in front of the blue car), convertible, Anton
3. minivan - third car in line, Diego
4.  white car, sedan, Ingrid
5. green car, immediately behind white car, Yuna, pickup


We know that Rachel is first, a blue car is second and the minivan is third.  We know that 3, 4 cannot be a convertible since one is a minivan and the other is a sedan.  We also know that a yellow car is in front of the convertible. 

Based off of that we know that car 5 cannot be a convertible since 4 is white and not yellow. So 1 or 2 is the convertible and SUV.  But we know that Rachel is in front of a blue car and the yellow car is in front of a convertible so that makes car #2 the blue convertible and car #1 the yellow SUV. 

Fill in what you do know and start to eliminate the outliers.  Hope that helps.

8 0
3 years ago
David bought 3 dvds and 4 books for $40 at a yard sale. Anna bought 1 dvd and 6 books for $18. How much did each dvd and book co
Ber [7]

Answer:

A book is worth $1 and a DVD is worth $12.

Step-by-step explanation:

The equations (2 unknowns and two equations, d is for a DVD and b is for a book):

For David: 3d+4b=40

For Anna: d+6b=18

Now multiply the second equation with -3 and add to the first equation:

3d+4b=40

−3d−18b=−54

Combined equation: −14b=−14 and b=1 (means that each book is worth $1).

 

Now for DVD price, use the second equation:

d=18−6 or d=12 (means that each DVD  is worth $12).

A book is worth $1 and a DVD is worth $12.

8 0
3 years ago
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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