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4 years ago

What is 5/6 x 24= plz tell me

1 answer:

7 0

The answer is 20 because you get .833333333 when you divide 5 and 6 then once you multiply it by 24 the answer comes out to 20

You might be interested in

Show that in a group of five people (where any two people are either friends or enemies), there are not necessarily three mutual

sveta [45]

Answer:

Lets call the people A,B,C,D and E. We need to find an example where there are not 3 mutual friends and not three mutual enemies.

Lets start by assuming that A has 2 friends, B and C, and 2 enemies, D and E.

Since we want A,B and C not to be mutually friends, then B and C neccesarily have to be enemies.

Now, since B and C are enemies, they cant be enemies at the same time with D, and they also cant be enemies at the same time with E. Therefore one of them has to be friends with D and one has to be friends with E.

Also, since D and E are enemies with A, they neccesarily need to be friends.

From the fact that D and E are friends, we coclude that they cant have a friend in common, as a consequence, B has to be friends with one of D and E and C has to be friend with the other. We can assume that B is friend with D and C is friend with E. If we use that, we have the following friendship configuration

---------------------

Friends of A: B, C

Enemies of A: D,E

-------------------------

Friends of B: A,D

Enemies of B: C, E

------------------------

Friends of C: A,E

Enemies of C: B,D

------------------------

Friends of D: B,E

Enemies of D: A,C

----------------------------

Friends of E: C,D

Enemies of E: A,B

--------------------------

As you can check, there is not three mutual friends nor three mutual enemies.

8 0

3 years ago

I can't figure out the answer

oksian1 [2.3K]

Answer:

63/100

Step-by-step explanation:

Im Pretty sure this the correct answer

8 0

3 years ago

Solve for x. 3x+11=9x-14

iVinArrow [24]

Answer:

x=25/6

Step-by-step explanation:

3x+11=9x-14

11=6x-14

25=6x

x=25/6

8 0

3 years ago

Read 2 more answers

a sixth grade class has 12 boys and 24 girls consider this statement: <br />for every two boys there are 4 girls do you ag

Nitella [24]

So the ratio of boys to girls is 12:24.. now, is that an equivalent ratio as 2 boys for 4 girls? or 2:4? let's see.

$\bf \cfrac{boys}{girls}\qquad 12:24\qquad \cfrac{12}{24}\implies \boxed{\cfrac{1}{2}}
\\\\\\
\cfrac{boys}{girls}\qquad 2:4\qquad \cfrac{2}{4}\implies \boxed{\cfrac{1}{2}}$

you could start simplifying the 12/24 fraction by dividing by some common multiple, say 2, and get 6/12, then divide by 3 and get 2/4 and stop there.

then you'll notice that 12/24 is really 2/4 in disguise.

4 0

3 years ago

Part 3 - Discussion/Explanation Question

SpyIntel [72]

Step-by-step explanation:

Vertical asymptote can be Identites if there is a factor only in the denominator. This means that the function will be infinitely discounted at that point.

For example,

$\frac{1}{x - 5}$

Set the expression in the denominator equal to 0, because you can't divide by 0.

$x - 5 = 0$

$x = 5$

So the vertical asymptote is x=5.

Disclaimer if you see something like this

$\frac{(x - 5)(x + 3)}{(x - 5)}$

x=5 won't be a vertical asymptote, it will be a hole because it in the numerator and denominator.

Horizontal:

If we have a function like this

$\frac{1}{x}$

We can determine what happens to the y values as x gets bigger, as x gets bigger, we will get smaller answers for y values. The y values will get closer to 0 but never reach it.

Remember a constant can be represent by

$a \times {x}^{0}$

For example,

$1 = 1 \times {x}^{0}$

$2 = 2 \times {x}^{0}$

And so on,

and

$x = {x}^{1}$

So our equation is basically

$\frac{1 \times {x}^{0} }{ {x}^{1} }$

Look at the degrees, since the numerator has a smaller degree than the denominator, the denominator will grow larger than the numerator as x gets larger, so since the larger number is the denominator, our y values will approach 0.

So anytime, the degree of the numerator < denominator, the horizontal asymptote is x=0.

Consider the function

$\frac{3 {x}^{2} }{ {x}^{2} + 1}$

As x get larger, the only thing that will matter will be the leading coefficient of the leading degree term. So as x approach infinity and negative infinity, the horizontal asymptote will the numerator of the leading coefficient/ the leading coefficient of the denominator

So in this case,

$x = \frac{3}{1}$

Finally, if the numerator has a greater degree than denominator, the value of horizontal asymptote will be larger and larger such there would be no horizontal asymptote instead of a oblique asymptote.

8 0

2 years ago