All categories

3 years ago

Which graph represents an odd function?

Mathematics

2 answers:

olga55 [171]3 years ago

7 0

The 3rd graph should be it

jeka57 [31]3 years ago

7 0

The third graph, with (-4,0),(-3,1),(-2,-3),(-1,-2), (1,2),(2,3),(3,-1),(4,0)

We see the pattern, if we have (x,f(x)) we have (-x, -f(x)). In other words

f(-x) = -f(x)

which is the definition of an odd function.

You might be interested in

Help me out pls!!!!!!

klio [65]

It is at (2,4) hope it helps

4 0

2 years ago

Read 2 more answers

Which line plot displays a data set with an outlier?<br><br> Plz answer quickly!

iren2701 [21]

The answer is the first data set because remember, an outlier is a number in a data set that is extremely far away from all of the other numbers.

You can easily tell which one is the outlier, the first data set's outlier is 13, because it is was placed so far away from the other data points.

We know that the first data set is correct because all of the other data sets have numbers that are clustered together.

~Hope I helped!~

7 0

3 years ago

What is the answer to this maths watch question? Look at screenshot attached. Thank you.

forsale [732]

14-x=9

14=x+9

x=14-9

x=5

14+5=19

x+y=19

x-y=(-5)

Adding both equations,

We get,

x+y=19

x-y=(-5)

2x=14

x=7

x+y=19

7+y=19

y=12

3 0

1 year ago

Read 2 more answers

How many bowls of cereal did he eat pls and thx

miv72 [106K]

36 bowls - since with one gallon he can eat 12 bowls (as demonstrated by 1/12), you multiply 12 by 3 to get 36 bowls.

8 0

3 years ago

Read 2 more answers

three bags of sweets weigh 27/4 kg. two of them have the same weight and the third bag is heavier than each of the bags of equal

Nana76 [90]

I'm here buddy,

so, let's take the value of the two bags with equal weight as x.

= x + x + (x + $\frac{6}{5}$ ) = $\frac{27}{4}$

= 3x + $\frac{6}{5}$ = $\frac{27}{4}$

= 3x = $\frac{27}{4}$ - $\frac{6}{5}$

( let's take the LCM of 4 and 5 = 20

= 3x = $\frac{135}{20}$ - $\frac{24}{20}$

= 3x = $\frac{111}{20}$

= x = $\frac{111}{20}$ ÷ $\frac{3}{1}$ = $\frac{111}{20}$ × $\frac{1}{3}$ = $\frac{37}{20}$

So, the weight of the equal bags are $\frac{37}{20}$ and the weight of the third bag ( heavy one ) is $\frac{37}{20}$ + $\frac{6}{5}$ = $\frac{37}{20}$ + $\frac{24}{20}$ = $\frac{61}{20}$

1st bag = $\frac{37}{20}$ kg

2nd bag = $\frac{37}{20}$ kg

3rd bag = $\frac{61}{20}$ kg

Hope it helps...

7 0

3 years ago