Could you explain a bit?
Weird; that's the question...
There are a maximum of 3*C(7,3) = 105 intersection points.
(a + (2n + 1)d)^2 - (a + (2n)d)^2 + ... + (a + d)^2 - a^2 = (a + (2n + 1) d + (a + 2nd + ... + a + d) + a = n(3a + (n + 2)d).
kn^2/lm = 15.
The answer is 34.
$a+b+c=750$
$a-30, b+30, c$
$a-30, b-20, c+50$
$a-30=b-20=c+50$
Solve.
$c=j+20$
$2j+20=100$
$j=40$
$c=60$
$4(40-t)=60$
Find $t$.
There is no diagram!!!
The answer is $\infty$.
To do this, use the code snippet:
if (4){
counter++;
}
The counter will always increase with whatever bounds you have set.
I had the same issue once too.
I tried to respond, but it just would cut off at the same place.
=^._.^=
We can write this as.
sqrt(16) < sqrt(n) < sqrt(100). Removing the sqrts.
16 < n < 100
so n is (16, 100)
This is the question I need help with idk why it wasn't showing up before
4 < sqrt(n) < 10
perhaps try responding to me rn.
lol. :))
Is something wrong with web calc? when I type in things and post it, it doesn't show up
Must have been A glitch or something since it says for me "Find the number of positive integers n so that 4
$P(G(a))=P(4+8a)=4+2\sqrt{6+8a}$.
We must have $8a+6\ge 0\implies 8a\ge -6\implies a\ge -\frac{3}{4}$. The minimum value of $a$ is $-\frac{3}{4}$, the maximum is non-existent.
Hmmmmmm the answer is a nonnegative integer, that is all I can say
I think you didn't finish the question.
All positive integers satisfy the statement "4".
(I'm not joking, write code that says )
Hey there!
CPhill actually did a nice solution to this problem a few years back! I hope he doesnt mind me linking his solution to this problem :)
https://web2.0calc.com/questions/find-all-cube-roots-of-8i
Hope this helped :)
You can find a circles circumference by multiplying its diameter (or 2*radius) by pi!!
You got this! Hope this helped :)
