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this way is probably not the most efficient way to do it, but i just listed out all the multiples of 8 that are between 1 and 100.

here is the list:

8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96.

when you count, there are 12 multiples of 8.

so the answer is 12/100, or 3/25. 

another method to solve this is to simply do 100 / 8, which is just 12 remainder 4.

since the quotient is 12, that means there are 12 multiples of 8 in between 1 and 100.

hope this helped! please let me know if you are confused about anything i did 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(x = {-(-5) \pm \sqrt{(-5)^2-4(1)(19)} \over 2(1)}\)

\(x = {5 \pm \sqrt{25-76} \over 2}\)

\(x = {5 \pm i\sqrt{51} \over 2}\)

a = \(\frac{5 + i\sqrt{51}}{2}\)

b = \(\frac{5 - i\sqrt{51}}{2}\)

(a - 1)(b - 1)

= \((\frac{5+i\sqrt{51}}{2} - 1)(\frac{5-i\sqrt{51}}{2} - 1)\)

= \((\frac{3+i\sqrt{51}}{2} )(\frac{3-i\sqrt{51}}{2} )\)

= \(\frac{(3+i\sqrt{51})(3-i\sqrt{51})}{4}\) 

= \(\frac{3^2-(i\sqrt{51})^2}{4}\) ==> difference in squares

= \(\frac{9 + 51}{4}\)

= \(\frac{60}{4}\)

= 15

(edit: I realized I made a mistake. Hopefully this answer is correct.)

No explanation is needed... I hope!?! 

check out the solutions hectictar and Melody have provided here:

Also,

If you want to add a useful answer to a locked question just send a message to a moderator, me or CPhill, and request it.

Don't lodge it as a formal 'complaint'

Will, please behave yourself.

You have answered this question 4 times now, each time with exactly the same answer.

Plus, I think your original link was a different one. 

Hey there, Guest!

So, for this equation, we'll need to use the Pythagorean Theorem. (It's okay if you don't know how to say that cause I don't either.)

\(a^2+b^2=c^2\)

So, basically:

\(a^2+24^2=25^2\)

And in this equation, 'a' would equal 7.

So the little extra part equals 7.

13+7=20

Bam, now we know that 20 is the length of our missing side.

Now for the easy part, adding.

20+25+24+13= 82

And now we've found our answer, easy as 1-2-3.

Hope this helped! :)

( ﾟдﾟ)つ Bye

ah okay, ty!!

Thank you again! I got $\frac{5}{12}$ and it was correct!

Graph it: https://www.desmos.com/calculator/9rbnk7rx1s

The trapezoid is a triangle + a rectangle. The probability is $\frac{\text{area of trapezoid}}{3*6}$

Make it a rectangle and a triangle and use Pythagorean theorem to find that the top part is $7$. The perimeter is $7+13+24+13+25=\boxed{82}$.

Forget about the girl/boy condition. We simply choose $2$ players from a group of 10, so the answer is $\binom{10}{2}=\boxed{45}$.

$x=-7$, so $49a-7b+c=0$.

No problem, keep the good work up! I hope to see you soon on the Alcumus hall of fame (or if it's a homework problem, on the forums).

While sam runs $3$ km, Ian runs $5+2=7$ km. So Ian is running $\frac{7}{3}$ as fast as Sam. Then $(s+4)=\frac{7}{3}s$. Now compute $s$ and finish.

Thank you for your solution! I finished the problem for 7 and 8 students, and got $\frac{11}{60}$, which was correct! Thanks again, this helped me a lot!

Thank you, this was what I needed!

Introduction to counting and probability is a great book ;)

Let's first calculate the probability for each club then average them.

5 student club - There are $\binom{5}{3}=10$ ways to choose the members. In $\binom{2}{2}\binom{3}{1}=3$ ways, the co-presidents are selected - the probability here is $\frac{3}{10}$.

7 student and 8 student clubs are analogous. Try it, I believe in you!

Noah rides at $n$ mph. Grady rides at $1.6n$ mph. Then $3.2n-2n=15\implies \boxed{n=12.5}$

$f(x)=y\implies 16x+3=y\implies x=\frac{y-3}{16}$ so $f^{-1}(x)=\frac{x-3}{16}$. The continuation is simple.

First, simplifying the right side gives $\frac{4x^2+5x+8x^2}{2x}=\frac{12x^2+5x}{2x}=\frac{12x+5}{2}$

Then, multiplying by $2(x+4)$ to get rid of fractions gives $2(3x+12)=(12x+5)(x+4)$.

Moving everything to one side gives $(12x+5)(x+4)-(6x+24)=0$.

Expanding everything gives $12x^2+53x+20-6x-24=0$.

Simplifying gives $12x^2+47x-4=0$.

Factoring this equation gives $(x-\frac{1}{12})(x+4)=0$.

So, our two solutions are $\frac{1}{12}$ and $-4$.

But wait!

$-4$ is extraneous, because plugging it in causes division by 0! That means $\boxed{x=\frac{1}{12}}$ is the only solution

The same question was asked (and answered) here:

Yep! Thanks again! Great solution, I forgot about the midpoint formula 

To have infinitely many solutions, we must have that the two equations are multiples of each other. Moving all variables to the left of the second equation gives $9a+12b=k$. Aha! The second equation is $3$ times the first! So we need $k$ to be $3(7)=\boxed{21}$