All categories

3 years ago

1/6 closest to 0, 1/2

Mathematics

2 answers:

allochka39001 [22]3 years ago

5 0

Closest to 0 my friend

elixir [45]3 years ago

3 0

1/6 is closest to zero.............

You might be interested in

You created a new playlist in 100 of your friends listened to it and shared if they likedThe new playlist or not nadhi said the

Nostrana [21]

Given

You created a new playlist in 100 of your friends listened to it and shared if they liked

The new playlist said the ratio Of people who liked the playlist to those who did not like the playlist is 75:25.

Dylan said that for every three people who like to playlist one person did not

Find out Do Nadhii and Dylan Agree or not.

To proof

As given in the question

created a new playlist in 100 of your friends listened to it and shared if they liked

new playlist said the ratio Of people who liked the playlist to those who did not like the playlist is 75:25.

Simply the ratio of people who liked the playlist to those who did not like the playlist

$= \frac{75}{25}$

$= \frac{3}{1}$

Thus

$\frac{People \ like \ the \ playlist}{People \ didnot \ like \ the\ playlist}=\frac{3}{1}$

also

Dylan said that for every three people who like to playlist one person did not

In the form of ratio written as

$\frac{People \ like \ the \ playlist}{People \ didnot \ like \ the\ playlist}=\frac{3}{1}$

the value of both new playlist and Dylan said about the ratio of people who liked the playlist to those who did not like the playlist is 3:1.

The value both of their ratio are equivalent .

Thus Nadhii and Dylan Agree

Hence proved

7 0

3 years ago

Help math question derivative!

atroni [7]

Let $f(x)=\sec^{-1}x$ . Then $\sec f(x)=x$ , and differentiating both sides with respect to $x$ gives

$(\sec f(x))'=\sec f(x)\tan f(x)\,f'(x)=1$
$f'(x)=\dfrac1{\sec f(x)\tan f(x)}$

Now, when $x=\sqrt2$ , you get

$(\sec^{-1})'(\sqrt2)=f'(\sqrt2)=\dfrac1{\sec\left(\sec^{-1}\sqrt2\right)\tan\left(\sec^{-1}\sqrt2\right)}$

You have $\sec^{-1}\sqrt2=\dfrac\pi4$ , so $\sec\left(\sec^{-1}\sqrt2\right)=\sqrt2$ and $\tan\left(\sec^{-1}\sqrt2\right)=1$ . So $(\sec^{-1})'(\sqrt2)=\dfrac1{\sqrt2\times1}=\dfrac1{\sqrt2}$

5 0

3 years ago

Quadrilateral GHIJ is similar to quadrilateral KLMN. Find the measure of side NK.

Leokris [45]

Answer:
NK = 14.4

Step-by-step explanation:

Given the following question:

We know that the two sides are similar
JI = 3, NM = 7.2

<u>The scale factor is 2.4 since....</u>

$3\times2.4=7.2$

Knowing the two sides are similar and if we multiply JI by 2.4, we get NM we can easily do the same for JG and NK.

JG = 6
$6\times2.4=14.4$
$=14.4$

NK = 14.4

Hope this helps.