All Answers 358089 Answers

Have you tried plotting A and C it on a grid and then trying to work it out from that?

2n^4−15/6n^4+17

\(\displaystyle \lim_{n\rightarrow \infty}\;\;\frac{2n^4−15}{6n^4+17}\\ \qquad \text{At an educated guess I would say 2/6 = 1/3}\\ \qquad \text{But I should be more mathematical about it}\\~\\ \text{divide through by n^4 top and bottom}\\~\\ \displaystyle \lim_{n\rightarrow \infty}\;\;\frac{2−\frac{15}{n^4}}{6+\frac{17}{n^4}}\\~\\ =\frac{2}{6}\\ =\frac{1}{3} \)

I first want to point out that the second example was incorrect becuase you need two 2's.

Honestly, I have no idea how to do this problem, Even Cphil is stuck on it: Proof at brainly if you copy paste this question

The axis of symmetry is x = 4.

To solve the problem, we can use the formula: distance = rate x time. Let's call the distance of the trip "d". The time it took to drive at 40 mph can be represented as t. The time it would have taken to drive at 30 mph can be represented as t + 20 minutes, or t + (20/60) hours.

From the first equation, we have: d = 40t

From the second equation, we have: d = 30(t + (20/60))

We can now set the two equations equal to each other:

40t = 30(t + (20/60)) Expanding the second equation: 40t = 30t + (30 * 20/60) Simplifying: 40t = 30t + 30

Solving for t: 10t = 30 t = 3

So the trip took 3 hours. Now we can use the first equation to find the distance:

d = 40t = 40 * 3 = 120 miles

Please proof read your questons and get rid of meaningles dollar signs

There are 9 stars (chocolate) and 7 - 1 = 6 bars. 

This makes for \({15 \choose 6} = 5005\) cases. 

If Dhruve gets exactly 2, there are now 7 stars, and 6 - 1 = 5 bars, which makes for \({12 \choose 5} = 792\) cases. 

So, the probability is \({792 \over 5005} = \color{brown}\boxed{72 \over 455}\)

It says it's a quiz... 

Maybe you aren't supposed to get help.

By the distance formula, the distance from the origin to the graph is \(\sqrt{(y-0)^2 + (x-0)^2} = \sqrt{y^2 + x^2}\)

We know that \(y = {1 \over 2}x^2 - 4\), meaning \({1 \over 2}x^2 = y + 4\), so \(x^2 = 2y + 8\)

Substituting this gives us \(\sqrt{y^2 + 2y + 8}\). 

The minimum value occurs at \(y = -{b \over 2a} \), which is -1.

This means the minimum is \(\sqrt{(-1)^2 + 2(-1) + 8} = \color{brown}\boxed{\sqrt{7}}\)

\(99 \times 100 \div 2 = \color{brown}\boxed{4950}\)

c) If ab = 0, either a or b = 0. 

If a = 0, b = -1, so \((a, b)\) is \(\color{brown}\boxed{(0, -1)}\)

If b = 0, a = 1, so \(\color{brown}\boxed{(1, 0)}\)

These are the only solutions that work

b) We already proved that \(a^2 + b^2 + ab = 1\).

Now, we write this as \((a+b)^2 - ab = 1\)

Because \(ab = 0\), we know \((a+b)^2 = 1\), meaning \(a + b = \color{brown}\boxed{-1, 1}\) 

The probability is 3/8.

The correct order is w + z, v + w, u + v, z + u.

The complete set is s, v, and w.

a) Note that \(a^3 - b^3 = (a-b)(a^2 + b^2 + ab)\)

Substituting what we have, we get \(1 = 1 (a^2 + b^2 + ab)\), which means that \(a^2 + b^2 + ab = 1\)

We can write \(a^2 + b^2 + ab \) as \((a - b)^2 + 3ab\). 

Using what we know, we have \(1^2 + 3ab = 1\), meaning \(ab = \color{brown}\boxed{0}\)

1.  Add the two equations to get \(2a^3=1352\),  so  \(a=26^{2/3}\)

2. Subtract the two equations to get \(6ab^2=6\)  so \(ab^2=1\)

Hence \(b^2=26^{-2/3}\)  and \(b=\pm26^{-1/3}\)

Take it from there!

ABCDEFG  that is 7 friends

Cut into 9 peices that is 8 bars.     

7+8=15

15C8=6435   ways to distribute.

If D gets 2 peices than you can take him and the 2 choc he is getting out  

then you are left with

ABCEFG   6 friends    divided into 7 parts that is 6 bars

6+6=12

12C6 = 924 possible ways

so that is    924/6435 =         28 / 195

I used the Stars and bars method to solve it. 

I could have make a logic error. You need to decide for yourelf whether to accept this answer.

Nice logic Imcool  

You got the right idea but got your permutations mixed up.

The boxes can have

0,0,0,4       3 possibilities............4! / 3! ==4

0,0,1,3       6 possibilities............4! / 2! ==12

0,0,2,2       3 possibilities............4! / 2!.2! ==6

0,1,1,2       3 possibilities.............4! / 2! ==12

1,1,1,1       1 possibilities..............4! / 4! ==1

Total ==4  +  12  +  6  +  12  +  1 ==35 ways - which is the same as:

[4 + 4 - 1] C 4 ==7 C 4 ==35 ways

How are: 45 and 810 "cozy", " if every digit in the number is either a 2 or next to a 2" ?? I think there is a mistake in your question. Check it carefully.

The balls are all the same so the only differenciating thing is how many balls are in each individual box.

The boxes can have

0,0,0,4       3 possibilities

0,0,1,3       6 possibilities

0,0,2,2       3 possibilities

0,1,1,2       3 possibilities

1,1,1,1       1 possibilities

So that are 3 + 6 + 3 + 3 + 1 = 16 unique outcomes

[12 C 10]  +  [12 C 11]  +  [12 C 12] ==[66 + 12 + 1]  ==79 sequences

Well lets do some algebra!

\(a^3 + 3ab^2 = 679\)

\(a^3 - 3ab^2 = 673\)

\(a^3 + 3ab^2 = 679 \) this is equal to

\(a^3 - 3ab^2 = 679 - 6ab^2\)

we can conclude that

\(673 = 679 - 6ab^2\)

\(6 = 6ab^2\)

\(ab^2 = 1\)

Most likely a will be very big because we need a^3 to equal something large since the second term is = to 3.

so a^3 = 676 so a = \(\sqrt[3]{676}\)

b = \(\sqrt{\frac{1}{\sqrt[3]{676}}},-\sqrt{\frac{1}{\sqrt[3]{676}}}\)

a - b = \(\sqrt[3]{676}-\frac{1}{\sqrt{\sqrt[3]{676}}},\sqrt[3]{676}+\frac{1}{\sqrt{\sqrt[3]{676}}}\) that is so definenly incorrect, so plz someone see if this is correct or not

Uhh its incorrect