All Answers 356328 Answers

Exactly my point.   Changing 5th to 4th makes it impossible.   I can show it's impossible but I'll leave you to see it for yourself.   I've solved and posted a solution for 5th already.

Well, I mean I just complete Pre-Calculus 2 days ago so probably.

I really didn't mean to ignore your questions.

I got the answer right though!!! My friend and I asked my teacher how to do it and she said that I was correct!!! YAY!

Thank you for your time!!! I really appreciate it!

This will explain better than I can:

https://brainly.in/question/6776543

We can use my favorite special formula for this:

U N D E F I N E D

This was answered here by MathProblemSolver:

https://web2.0calc.com/questions/counting_135

The immage below is simmilar question...The different between the two question > "5k is a perfect 4th power." and your question "5k is a perfect 5th power."

Ref:  https://web2.0calc.com/questions/plz-help-fast-urgent

First find out the exact formula for $P(x)$.   To do that, you can use the method as described in the solution at https://web2.0calc.com/questions/hard-algebra_17#r1.   Once the formula for $P(x)$ is known, just plug in 1 for x to get the answer.
 

I think no such k exists.

$\frac{2x+3}{4} = \frac{x+3}{2}$

$2x + 3 = 2x + 6$

$0 = -3$
$\boxed{\varnothing}$

Let the two boys and Jill be one: There are $\binom{3}{2} = 3$ possibilities to permute them.

Then there are $5-3 + 1 = 3$ people. There is $\binom{5}{3} = 10.$

the answer is $10\cdot 3 = \boxed{30}$

54000 has the prime divisors of 2, 3, and 5. 

So the number can use 2, 3, or 5 once. 

2

3

5

2*3

2*5

3*5

2*3*5

1, 2, 3, 5, 6, 10, 15, 30 are the positive square free divisors, and they sum up to 72. :)

=^._.^=

 (a-bi)^2 = 8 - 6i + 13 - 4i

a^2 - b^2 - 2abi = 21 - 10i

-2abi = -10i

ab = 5

a^2 - b^2 = 21

Can you take it from here?

=^._.^=

LCM (2, 3, 5, 7, 19) = 3990

3990, 3990*2

It stops there since 3990*3 is not a 4 digit number. 

=^._.^=

g(2x) = 2(g(x))

(2x + 3)/4 = 2((x + 3)/4)

(2x + 3)/4 = (x + 3)/2

2(2x + 3) = 4(x+3)

4x + 6 = 4x + 12

6 = 12???

I don't think this is possible.

=^._>^=

Nice, I had the same idea but I thought that 0 was a very suspicious answer. 

=^._.^=

Try to write $1/a^2+1/b^2$ as an expression involving only $a+b$ and $ab$.  The values of $a+b$ and $ab$ can be obtained from the coefficients of $5x^2-11x+4$ almost by inspection. (Vieta)

Note that $(m + 1/m)^3 = m^3 + 3m + 3/m + 1/m^3 = (m^3 + 1/m^3) + 3(m+1/m)$.  So $m^3+1/m^3 = (m+1/m)^3 - 3(m+1/m) =8^3-3*8$

First of all, ABCD has area of 32.  

Knowing that P is directly above A (from PA perpendicular to AD and AB), we use the Pythagorean Theorem on triangle PAB, getting PA = \(\sqrt297\).  

With this, we have the volume of PABCD as \(\frac{32*\sqrt{297}}{3}\), or \(32\sqrt{33}\)

To find divisors, we must prime factorize the number.  

1984*2 = 2^7 × 31

Using the divisor formula, however, we find that there is no integer value of k that this is true (8*2*(any integer) cannot = 21)

Because of this, there is no number that satisfies this, meaning the sum is 0

Additional material on the divisor formula I found on google is here: https://www.themathdoctors.org/counting-divisors-of-a-number/

Sidenote:  If there was no extra *2, then k=31 would make the number 1984*31 have 21 divisors.  

Repeating.  

While I have little experience with fractions within bases, the base converter calculator expressed 228/47 as 4.A2677AB8B29851339B5A574A2677AB8B29851339B5A574A0

With the proper highlights, we can find where it repeats

4.A2677AB8B29851339B5A574A2677AB8B29851339B5A574A2677AB8B29851339B5A574A2677AB8B29851339B5A57, and so on.  

Yeah, I know this was cheating, but at least it will help.  :)

The x-intercept occurs when y=0.  x-intercept is (6, 0)

The y-intercept occurs when x=0.  y-intercept is (0, -3)

Applying the distance formula/Pythagorean theorem, we have the distance between the points to be \(\sqrt{9+36}\), or \(\sqrt{45}\), equal to \({3}\sqrt{5}\)

2/5  j = 3/4 k         or     k = 8/15 j 

    k already has   LESS money than j            giving him any amount of money will not make them equal

The two expressions must be equal when x = 3, so:

3 + 2 = 2*3^2 + 3*3 + a

so

5 = 27 + a

I'll leave you to find a from this.

Subtract x from both sides and add 5 to both sides.  What do you get?

Rounding 345.67 to the nearest

thousand:   0

hundred:     300

ten:             340

unit:            346

tenth:          345.7

hundredth:  345.67

You can select the largest!