Last week, he completed 2/5 of the levels. This week he completes the remaining amount of games times 1/3, which is (3/5*1/3)=1/5. Therefore, he has 2/5 of the total levels remaining. If g is the total games, we have that 2/5*g=24, so g=24*5/2=60 total games.
n==119, 238, 357 ==3 such values of n
Okay, for a second I forgot that negative numbers could be squared to get a positive number. This thread is useless now, sorry for making it...
What is the ratio of BY to YA?
Hello Guest!
\(BY:YA=\frac{12}{\sqrt{2}}:(18\cdot \sqrt{2}-\frac{12}{\sqrt{2}})\\ BY:YA=12:(18\cdot 2-12)=12:24\\ \color{blue}BY:YA=1:2 \)
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Hi Melody: Did you miss 256 ==6 in the last category "All different" ?
Here's an alternative approach:
do the substitution, get the values then go to a factor calculator to get any common factors.
You can use the web2.0calculator
If you want the facors of 20 say, you'd put in
factor(20) and press =
{2, 3, 5, 5, 5, 6, 6, 6}
all the same
666 1
555 1
2 the same
665 3
663 3
662 3
556 3
553 3
552 3
All different
235 3!=6
236 3!=6
356 3!=6
2+18+18 = 38
What did you get?
I would be interested to see someones approach as well
Thanks i got it
x=t-3
so y=x+3
Substitute that into the second equation and you have your non-parametric form
Tidy it up for into standard form.
Do that and come back with what you have done.
If you want help how about presenting your question properly.
Put the stuff between the dollar signs into a LaTex box.
Just by visual inspection.
If x=8 = 8mod 9
If x=8 = 2 mod 6
x=18k +8
I am not sure but answer may be 8
sales funnel expert / consultant for your online business
Rectangle A 8 * 8 = 64 mi^2
Rectangle B
6 * 4 = 24 mi^2
Triangle C
1/2 * 4 * 5 = 10 mi^2
64 mi^2 + 24 mi^2 + 10 mi^2 = 98 mi^2
CE = sqrt(5).
XY = 36.
\(a_1 = 5\)
I'm more concerned about how you actually get there. I need to write a proof on how to factor it.
My calculator says
a*b^3 - a^3*b + b*c^3 - b^3*c + c*a^3 - c^3*a = (a - b)(a + c)(b + c)(a + b - c)
The soution is \(n \equiv 8 \pmod{54}\)
By partial fractions, (A,B) = (4/3,-1/3).
XD